Coordinate Geometry

Contents

**Describe the Cartesian and Polar Coordinates of a Point** 1

**Find the relation between the Polar and Cartesian Coordinates** 2

**Find the distance between two Coordinate Points** 3

**Describe the Internal division of a Line** 5

**Find the coordinate of a point which divides internally a line segment in m:n** 6

**Describe the External division of a Line** 7

**Find the coordinate of a point which divides externally a line segment in the ratio m : n** 8

**Describe the Centroid of a Triangle** 9

**Find the Coordinates of the Centroid of a Triangle in terms of vertices** 10

**Describe the Incentre of a Triangle** 11

**Find the Coordinates of Incentre of a Triangle in terms of vertices and sides** 13

**Find the Area of a Triangle whose vertices are known (Determinant method)** 14

**Describe the Locus and its Equation** 15

**Recall Three-dimensional Geometry** 17

**Describe the distance formulation in 3D** 18

**Find the Direction Ratios if its Direction Cosines are known** 19

**Describe Direction Ratios and Direction Cosines of a Line** 19

**Describe the Equation of a Line through a given point and parallel to the given Vector** 19

**Describe the Equation of a Line passing through two given Points** 19

**Recall the Angle between two Lines** 19

**Recall the distance between two Skew Lines and two Parallel Lines** 19

**Describe the Angle between two Planes** 19

**Describe the Angle between a Line and a Plane** 19

**Describe the Coplanarity of two lines** 19

**Recall the distance formulae of a Point from a Plane** 19

**Describe the equation of a Circle** 19

**Find the center and radius of a given Circle** 19

**Describe the diametric form of a Circle** 19

**Find the equation of a Circle passing through three non-collinear points** 19

**Describe the intercepts made on the axes by a Circle** 19

**Find the equation of a Circle when touches or cut the axes** 19

**Describe the concept of Conic Section** 19

**Recall the term Eccentricity** 19

**Recall the nature of Conics in view of Second Degree Equation** 19

**Describe the concept of Parabola** 19

**Describe all standard forms of a Parabola** 19

**Describe the standard equation of an Ellipse** 19

**Recall the terms of the Ellipse: Major Axis, Minor Axis, and Latus Rectum** 19

**Recall the Basic Terminologies used in the Hyperbola** 19

**Recall standard equation of Hyperbola** 19

**Describe the Cartesian and Polar Coordinates of a Point**

In geometry, a point is an exact location or position represented by a dot. A point can be defined using different coordinate systems, such as Cartesian coordinates and polar coordinates.

Cartesian coordinates, also known as rectangular coordinates, are defined using the x and y axes. The x-axis is the horizontal axis, and the y-axis is the vertical axis. The point of intersection between the x and y axes is called the origin. To locate a point on the plane, its distance from the origin along the x-axis and y-axis is measured and represented as (x, y) in the form of an ordered pair.

For example, let us consider a point P, which is 3 units to the right of the origin and 4 units above the origin. The coordinates of point P in the Cartesian coordinate system are (3,4).

Polar coordinates, on the other hand, represent a point using its distance from the origin and the angle it makes with the positive x-axis. The distance of a point from the origin is called the radial distance, and the angle made by the point with the positive x-axis is called the angular distance or polar angle. A point in polar coordinates is represented as (r,θ), where r is the radial distance and θ is the polar angle in radians.

For example, let us consider a point Q, which is 5 units away from the origin and makes an angle of 60 degrees with the positive x-axis. The polar coordinates of point Q are (5,π/3) or (5,60°).

It is important to note that the same point can be represented using different coordinate systems. For example, the point P mentioned earlier can also be represented using polar coordinates as (5, arctan(4/3)).

**Find the relation between the Polar and Cartesian Coordinates**

The polar and Cartesian coordinates are two different ways of representing the position of a point in a two-dimensional plane. The Cartesian coordinate system uses two perpendicular axes, usually labelled as the x-axis and y-axis, to specify the location of a point by its horizontal and vertical distances from the origin. The horizontal distance is measured along the x-axis, and the vertical distance is measured along the y-axis.

On the other hand, the polar coordinate system uses a different approach to represent the position of a point. In this system, a point is represented by an angle and a distance from the origin. The angle, usually denoted as θ (θ), is measured in a counterclockwise direction from the positive x-axis. The distance, denoted as r, is the straight-line distance from the point to the origin.

The relationship between the Cartesian and polar coordinates can be expressed mathematically as follows:

x = r *cos(θ)*

*y = r *sin(θ)

where x and y are the Cartesian coordinates of the point, and r and θ are the polar coordinates.

Conversely, the polar coordinates can be found from the Cartesian coordinates using the following equations:

r = sqrt(x^{2} + y^{2})

θ = arctan(y/x)

where sqrt denotes the square root and arctan denotes the inverse tangent function.

For example, consider a point P with Cartesian coordinates (3, 4). We can find its polar coordinates as follows:

r = sqrt(3^{2 }+ 4^{2}) = 5

θ = arctan(4/3) ≈ 0.93 radians (or 53.13 degrees)

Therefore, the polar coordinates of the point P are (5, 0.93). Similarly, we can find the Cartesian coordinates of a point given its polar coordinates.

**Find the distance between two Coordinate Points**

In coordinate geometry, one of the essential operations is to calculate the distance between two points. The distance formula helps to find the straight-line distance between any two points on the coordinate plane. The distance formula is derived from the Pythagorean theorem, which is applicable to right-angled triangles.

Consider two points on a plane A(x1, y1) and B(x2, y2). The distance between these two points is given by the formula:

d = √[(x^{2} – x^{1})^{2} + (y^{2} – y^{1})^{2}]

Where d is the distance between the two points.

Let’s take an example to understand this formula better. Suppose we have two points on a plane, A(2, 3) and B(5, 7). We need to calculate the distance between these two points.

Using the distance formula, we get:

d = √[(5 – 2)^{2} + (7 – 3)^{2}]

= √[3^{2} + 4^{2}]

= √(9 + 16)

= √25

= 5

Therefore, the distance between the points A(2, 3) and B(5, 7) is 5 units.

Another way to find the distance between two points is by using the Pythagorean theorem, which states that the sum of the squares of the legs of a right-angled triangle is equal to the square of the hypotenuse. We can form a right-angled triangle with the two given points and then use the Pythagorean theorem to find the distance.

Let’s take the same example as before, A(2, 3) and B(5, 7). We can form a right-angled triangle by drawing a perpendicular line from point B to the x-axis. The length of this line is (y_{2} – y_{1}) = (7 – 3) = 4 units. The length of the base of this triangle is (x_{2} – x_{1}) = (5 – 2) = 3 units. Therefore, the length of the hypotenuse of this right-angled triangle, which is the distance between points A and B, is given by the Pythagorean theorem as:

c^{2} = a^{2 }+ b^{2}

c^{2} = 3^{2} + 4^{2}

c^{2} = 9 + 16

c^{2 }= 25

c = √25

c = 5

So, the distance between the points A(2, 3) and B(5, 7) is 5 units. ** **

**Describe the Internal division of a Line**

In geometry, a line is an infinitely long, straight path with no width or height. When a line segment is divided into two parts by a point on it, the two parts are called the line segments’ division. The ratio of the two parts of the line segment is known as its division ratio. This ratio is also called the section formula, which is used to calculate the point’s coordinates that divide the line into the given ratio.

Let’s suppose we have a line segment AB with two endpoints A (x_{1}, y_{1}) and B (x_{2}, y_{2}), and a point P (x, y) divides the line into m:n ratio, where m+n = k, then:

- The coordinates of the point P can be calculated using the section formula as:

x = (mx_{2} + nx_{1})/k and y = (my_{2} + ny_{1})/k

- The distance of the point P from A can be calculated using the distance formula as:

AP = sqrt((x – x_{1})^{2} + (y – y_{1})^{2})

- The distance of the point P from B can be calculated using the distance formula as:

BP = sqrt((x – x_{2})^{2} + (y – y_{2})^{2})

The internal division of a line segment is when the point P divides the line segment AB internally such that it lies between A and B. The ratio m:n can be negative or positive, but the sum of the ratios must always be a constant value k.

For example, suppose we have a line segment AB with endpoints A(2, 3) and B(6, 9), and a point P divides the line segment into 2:3 ratio internally. Then the coordinates of the point P can be calculated as:

x = (3*6 + 2*2)/5 = 4.4

y = (3*9 + 2*3)/5 = 6.6

Therefore, the coordinates of point P are (4.4, 6.6), and the distance of point P from A is:

AP = sqrt((4.4 – 2)^{2} + (6.6 – 3)^{2}) = sqrt(20) = 4.47

Similarly, the distance of point P from B is:

BP = sqrt((4.4 – 6)^{2} + (6.6 – 9)^{2}) = sqrt(20) = 4.47

Hence, the point P divides the line segment AB internally in the ratio of 2:3.

**Find the coordinate of a point which divides internally a line segment in m:n**

Suppose we have a line segment AB, and we want to find the coordinates of a point P that divides AB in the ratio m:n, where m and n are positive integers.

Let the coordinates of point A be (x1, y1) and the coordinates of point B be (x_{2}, y2). Then, the coordinates of point P can be found using the following formula:

x-coordinate of P = [(nx_{1}) + (mx_{2})] / (m + n)

y-coordinate of P = [(ny_{1}) + (my_{2})] / (m + n)

Therefore, the coordinates of point P are:

P = ( [(nx_{1}) + (mx_{2})] / (m + n) , [(ny_{1}) + (my_{2})] / (m + n) )

Example

Let’s say we have a line segment AB with endpoints A(2, 4) and B(8, 10), and we want to find the coordinates of a point P that divides AB in the ratio 3:4.

Using the formula I provided earlier, the coordinates of P are:

x-coordinate of P = [(42) + (38)] / (3 + 4) = 4.57 (rounded to two decimal places)

y-coordinate of P = [(44) + (310)] / (3 + 4) = 6.86 (rounded to two decimal places)

Therefore, the coordinates of point P, which divides the line segment AB in the ratio 3:4, are approximately P(4.57, 6.86).

**Describe the External division of a Line**

External division of a line refers to the division of a line segment into two parts by a point that is located outside the line segment. This point is called the external point of division.

To determine the external point of division, we need to extend one of the lines beyond its endpoint to locate the external point. Then, we divide the line segment into two parts in such a way that the ratio of the length of the smaller part to the length of the larger part is equal to a given ratio.

Suppose we have a line segment AB with endpoints A and B, and we want to find an external point P that divides AB in the ratio m:n. To find the coordinates of point P, we use the following formula:

x-coordinate of P = [(nx_{1}) – (mx_{2})] / (n – m)

y-coordinate of P = [(ny_{1}) – (my_{2})] / (n – m)

where (x_{1}, y_{1}) and (x_{2}, y_{2}) are the coordinates of points A and B, respectively.

Note that if m > n, then the external point of division will lie on the extension of AB beyond point A. On the other hand, if n > m, then the external point of division will lie on the extension of AB beyond point B.

Example

Let’s say we have a line segment AB with endpoints A(3, 5) and B(8, 10), and we want to find an external point P that divides AB in the ratio 3:2.

To find the coordinates of point P, we use the following formula:

x-coordinate of P = [(23) – (38)] / (2 – 3) = -5

y-coordinate of P = [(25) – (310)] / (2 – 3) = -5

Therefore, the external point P with coordinates (-5, -5) divides the line segment AB in the ratio 3:2. Note that this point lies on the extension of the line AB beyond point A, since the smaller part of the segment is the one adjacent to point A.

**Find the coordinate of a point which divides externally a line segment in the ratio m : n**

External division of a line segment occurs when the point dividing the line segment is outside the line segment. The point divides the line segment in a certain ratio. Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be the end-points of the line segment, and P(x, y) be the point which divides AB externally in the ratio m : n.

We can find the coordinates of point P by using the section formula as follows:

Let the coordinates of point P be (x, y).

Then we have:

x = (m * x_{2 }+ n * x_{1}) / (m + n)

y = (m * y_{2 }+ n * y_{1}) / (m + n)

For example, suppose we want to find the coordinates of a point P that divides the line segment AB with endpoints A(2, 4) and B(6, 8) externally in the ratio 2:3. Using the external section formula, we get:

x = (3 * 2 + 2 * 6) / (3 + 2) = 16/5

y = (3 * 4 + 2 * 8) / (3 + 2) = 28/5

Therefore, the coordinates of the point P are (16/5, 28/5).

**Describe the Centroid of a Triangle**

The centroid is a point of concurrence of the medians of a triangle. A median of a triangle is a line segment that connects a vertex of a triangle to the midpoint of the opposite side. The centroid divides each median in the ratio of 2:1.

The coordinates of the centroid of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2)}, and (x_{3}, y_{3}) are given by:

(x, y) = ((x_{1} + x_{2 }+ x3)/3, (y_{1} + y_{2 }+ y_{3})/3)

This means that the x-coordinate of the centroid is the average of the x-coordinates of the vertices, and the y-coordinate of the centroid is the average of the y-coordinates of the vertices.

The centroid is often referred to as the center of gravity of the triangle because if the triangle is cut out of a piece of cardboard and the centroid is supported, the triangle will balance perfectly on the point of the centroid.

Example:

Consider a triangle with vertices A(1, 4), B(4, 8), and C(6, 2). We can find the centroid of the triangle by averaging the x-coordinates and y-coordinates of the vertices:

x = (1 + 4 + 6)/3 = 11/3

y = (4 + 8 + 2)/3 = 14/3

So the centroid of the triangle is (11/3, 14/3).

**Find the Coordinates of the Centroid of a Triangle in terms of vertices**

The centroid is the point where the three medians of a triangle intersect. The median is a line segment drawn from one vertex of a triangle to the midpoint of the opposite side. The centroid divides each median into two parts, with the ratio of the lengths of the two parts being 2:1.

Let’s say we have a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}). The coordinates of the centroid (G) of the triangle can be found using the following formulas:

xG = (x_{1} + x_{2} + x_{3}) / 3

yG = (y_{1} + y_{2} + y_{3}) / 3

For example, let’s say we have a triangle with vertices (2, 3), (5, 7), and (9, 4). We can find the centroid of the triangle as follows:

xG = (2 + 5 + 9) / 3 = 16 / 3

yG = (3 + 7 + 4) / 3 = 14 / 3

Therefore, the centroid of the triangle is (16/3, 14/3).

The centroid of a triangle has several interesting properties. For example, it is the center of mass of the triangle, and if the triangle is made of a thin, uniform material, it will balance perfectly on a point located at the centroid. Additionally, the distance from the centroid to each vertex of the triangle is two-thirds of the length of the median connecting the vertex to the opposite side.

**Describe the Incentre of a Triangle**

The incenter of a triangle is the center of the circle inscribed inside the triangle. The incenter is the intersection of the angle bisectors of the triangle, and it is equidistant from each side of the triangle. The circle inscribed in the triangle touches each side of the triangle at a point called the point of contact.

The incenter of a triangle has several important properties:

- The incenter is the center of the circle inscribed inside the triangle.
- The distance from the incenter to each side of the triangle is equal.
- The incenter is equidistant from the three sides of the triangle.
- The angle bisectors at the incenter intersect to form an equilateral triangle.

To find the coordinates of the incenter of a triangle, we can use the following formulae:

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) be the vertices of the triangle. Then the coordinates of the incenter I are given by:

Ix = (ax * x_{1} + bx * x_{2} + cx * x_{3}) / (ax + bx + cx)

Iy = (ay * y_{1 }+ by * y_{2} + cy * y_{3}) / (ay + by + cy)

where

ax = sqrt((y_{2} – y_{3})^{2} + (x_{2} – x_{3})^{2})

ay = sqrt((x_{2} – x_{3})^{2} + (y_{2} – y_{3})^{2})

bx = sqrt((y_{3} – y_{1})^{2} + (x_{3} – x_{1})^{2})

by = sqrt((x_{3 }– x_{1})^{2} + (y_{3} – y_{1})^{2})

cx = sqrt((y_{1 }– y_{2})^{2} + (x_{1} – x_{2})^{2})

cy = sqrt((x_{1} – x_{2})^{2} + (y_{1} – y_{2})^{2})

For example, let’s find the coordinates of the incenter of the triangle with vertices A(0,0), B(4,0), and C(0,3).

ax = sqrt((0 – 3)^{2} + (4 – 0)^{2}) = 5

ay = sqrt((4 – 0)^{2} + (0 – 3)^{2}) = 5

bx = sqrt((3 – 0)^{2} + (0 – 0)^{2}) = 3

by = sqrt((0 – 0)^{2} + (0 – 3)^{2}) = 3

cx = sqrt((0 – 4)^{2} + (0 – 0)^{2}) = 4

cy = sqrt((0 – 0)^{2} + (0 – 0)^{2}) = 0

Ix = (ax * x_{1 }+ bx * x_{2} + cx * x_{3}) / (ax + bx + cx) = (5*0 + 3*4 + 4*0) / (5 + 3 + 4) = 1.33*

*Iy = (ay * y _{1 }+ by * y_{2} + cy * y_{3}) / (ay + by + cy) = (5*0 + 3

*0 + 0*3) / (5 + 3 + 4) = 0.46

Therefore, the coordinates of the incenter I are (1.33, 0.46).

**Find the Coordinates of Incentre of a Triangle in terms of vertices and sides**

The Incentre is the point of intersection of the angle bisectors of a triangle. It is the centre of the incircle which is the circle that touches all three sides of the triangle. The Incentre is equidistant from all three sides of the triangle.

To find the coordinates of the Incentre, we can use the following formula:

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) be the vertices of a triangle. Let a, b, and c be the lengths of the sides opposite to the vertices A, B, and C, respectively. The coordinates of the Incentre I are given by:

Ix = (a x_{1 }+ b x_{2} + c x_{3}) / (a + b + c)

Iy = (a y_{1 }+ b y_{2} + c y_{3}) / (a + b + c)

Example:

Let us find the coordinates of the Incentre of a triangle with vertices A(0,0), B(4,0), and C(0,3).

We first find the lengths of the sides of the triangle.

a = BC = sqrt((3-0)^{2} + (0-4)^{2}) = 5

b = AC = sqrt((3-0)^{2} + (0-0)^{2}) = 3

c = AB = sqrt((0-0)^{2} + (4-0)^{2}) = 4

Using the formula, we can find the coordinates of the Incentre I:

Ix = (a x_{1} + b x_{2} + c x_{3}) / (a + b + c) = (5*0 + 3*4 + 4*0) / (5+3+4) = 1.2*

*Iy = (a y _{1 }+ b y_{2} + c y_{3}) / (a + b + c) = (5*0 + 3

*0 + 4*3) / (5+3+4) = 1.4

Therefore, the coordinates of the Incentre I are (1.2, 1.4).

**Find the Area of a Triangle whose vertices are known (Determinant method)**

The determinant method is one of the most efficient ways of finding the area of a triangle when the coordinates of its vertices are given. In this method, the determinant of the matrix formed by the coordinates of the vertices of the triangle is calculated, and then the absolute value of the determinant is divided by 2 to get the area of the triangle.

Example:

Let’s consider a triangle with the vertices A(1, 2), B(4, 5), and C(7, 8). To find the area of this triangle using the determinant method, we can follow the following steps:

Step 1: Write the coordinates of the vertices in the form of a matrix

Step 2: Calculate the determinant of the matrix A. We can do this by expanding along the first row or column:

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| 1 2 | | 4 5 | = (1*5 – 2*4) – (1*8 – 2*7) + (4*8 – 5*7) = -3 | 7 8 |

Step 3: Take the absolute value of the determinant: |det(A)| = |-3| = 3.

Step 4: Divide the absolute value of the determinant by 2 to get the area of the triangle:

Area = |det(A)| / 2 = 3 / 2 = 1.5 square units.

Therefore, the area of the triangle with vertices A(1, 2), B(4, 5), and C(7, 8) is 1.5 square units.

Conclusion:

The determinant method is a simple and efficient way to find the area of a triangle when the coordinates of its vertices are known. It works for all types of triangles, including acute, obtuse, and right triangles.

**Describe the Locus and its Equation**

A locus is a set of points that satisfy a particular condition or set of conditions. In geometry, it is the path traced by a moving point that satisfies certain conditions. The equation of a locus describes the relationship between the coordinates of the points that lie on the locus. In this topic, we will discuss the concept of locus and its equation with suitable examples.

Example:

Let’s consider a point P(x, y) that moves in a plane, such that its distance from the point A(2, 3) is always equal to its distance from the point B(4, 5). The locus of the point P is a line that passes through the midpoint of the line segment AB.

To find the equation of the locus, we can follow the following steps:

Step 1: Let M be the midpoint of the line segment AB, which is given by:

M = ((2+4)/2, (3+5)/2) = (3, 4)

Step 2: Let P(x, y) be any point on the locus. Then, the distance of P from A is given by:

PA = sqrt((x-2)^{2} + (y-3)^{2})

Step 3: Similarly, the distance of P from B is given by:

PB = sqrt((x-4)^{2} + (y-5)^{2})

Step 4: Since PA = PB, we can write:

sqrt((x-2)^{2} + (y-3)^{2}) = sqrt((x-4)^{2} + (y-5)^{2})

Squaring both sides, we get:

(x-2)^{2} + (y-3)^{2} = (x-4)^{2} + (y-5)^{2}

Expanding the squares and simplifying, we get:

-2x + 6y = 2

This is the equation of the locus of the point P.

Conclusion:

The locus of a point is a set of points that satisfy a particular condition or set of conditions. The equation of a locus describes the relationship between the coordinates of the points that lie on the locus. The concept of locus and its equation are important in geometry and have numerous applications in various fields of mathematics, physics, and engineering.

**Recall Three-dimensional Geometry**

Three-dimensional geometry, also known as 3D geometry, deals with the study of objects in three-dimensional space. It is an essential branch of mathematics that has numerous applications in various fields such as physics, engineering, and computer graphics. In this topic, we will recall the basics of three-dimensional geometry.

Example:

Let’s consider a point P in three-dimensional space. The position of the point P is described by its coordinates (x, y, z), where x, y, and z are the distances of the point P from the three perpendicular coordinate axes, respectively. The three coordinate axes are denoted by x-axis, y-axis, and z-axis.

The three-dimensional space can be represented by a Cartesian coordinate system, which consists of three perpendicular coordinate axes. The point of intersection of the three axes is called the origin, and it is denoted by O. The positive direction of the x-axis is taken to be towards the right, the positive direction of the y-axis is taken to be upwards, and the positive direction of the z-axis is taken to be towards the viewer.

In three-dimensional space, we can define various geometric objects such as points, lines, planes, and spheres. For example, a line in three-dimensional space can be defined by two points or a point and a direction vector. A plane in three-dimensional space can be defined by a point and a normal vector. A sphere in three-dimensional space can be defined by a center point and a radius.

Three-dimensional geometry also involves various concepts such as distance, angle, and projection. For example, the distance between two points in three-dimensional space is given by the formula:

d = sqrt((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2})

The angle between two lines in three-dimensional space is given by the formula:

cos(θ) = (a.b) / (|a| |b|)

where a and b are the direction vectors of the two lines.

The projection of a point on a plane in three-dimensional space is given by the formula:

P = Q + t(n)

where Q is the point to be projected, n is the normal vector of the plane, and t is a scalar.

Conclusion:

Three-dimensional geometry is an essential branch of mathematics that deals with the study of objects in three-dimensional space. It involves various geometric objects, concepts, and formulas that have numerous applications in various fields such as physics, engineering, and computer graphics. A good understanding of three-dimensional geometry is essential for solving problems in these fields.

**Describe the distance formulation in 3D**

In three-dimensional geometry, the distance between two points is a fundamental concept that is used in many applications, such as determining the distance between two objects, finding the shortest path between two points, and measuring the length of a line segment. In this topic, we will describe the distance formulation in 3D.

Formula for the Distance between Two Points in 3D:

The distance between two points P1(x1, y1, z1) and P2(x2, y2, z2) in 3D space is given by the following formula:

distance = sqrt((x2 – x1)^{2} + (y2 – y1)^{2} + (z2 – z1)^{2})

The distance formula in 3D is an extension of the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Example:

Let’s consider two points, P1(1, 2, 3) and P2(4, 5, 6), in 3D space. The distance between these two points can be found using the distance formula as follows:

distance = sqrt((4 – 1)^{2} + (5 – 2)^{2} + (6 – 3)^{2})

distance = sqrt(3^{2} + 3^{2} + 3^{2})

distance = sqrt(27)

distance = 3 sqrt(3)

Therefore, the distance between the two points P1 and P2 in 3D space is 3sqrt(3) units.

Conclusion:

The distance formulation in 3D is a fundamental concept that is used in many applications of three-dimensional geometry. The formula for the distance between two points in 3D space is an extension of the Pythagorean theorem, and it can be used to find the distance between any two points in 3D space. A good understanding of the distance formulation in 3D is essential for solving problems that involve finding the distance between two objects or the shortest path between two points in three-dimensional space.

**Find the Direction Ratios if its Direction Cosines are known**

In three-dimensional geometry, the direction ratios and direction cosines of a line are two important concepts used to determine the direction of a line. Direction ratios are used to determine the direction of a line by expressing the ratios of the differences of the coordinates of two points on the line in a suitable form. Direction cosines are used to express the direction of a line in terms of the angles it makes with the positive x, y, and z-axes. In this topic, we will learn how to find the direction ratios of a line if its direction cosines are known.

Formula for Finding Direction Ratios:

If l, m, and n are the direction cosines of a line, then the direction ratios of the line are given by the following formula:

Direction ratio of x = l/k

Direction ratio of y = m/k

Direction ratio of z = n/k

where k = sqrt(l^{2 }+ m^{2} + n^{2})

Example:

Let’s consider a line with direction cosines l = 1/3, m = -2/3, and n = 2/3. To find the direction ratios of the line, we use the formula mentioned above.

k = sqrt(l^{2} + m^{2} + n^{2}) = sqrt((1/3)^{2} + (-2/3)^{2} + (2/3)^{2}) = sqrt(1/3)

Therefore, the direction ratios of the line are:

Direction ratio of x = (1/3) / sqrt(1/3) = 1/sqrt(3)

Direction ratio of y = (-2/3) / sqrt(1/3) = -2/sqrt(3)

Direction ratio of z = (2/3) / sqrt(1/3) = 2/sqrt(3)

Conclusion:

The direction ratios of a line can be found if its direction cosines are known using the formula mentioned above. A good understanding of the concept of direction ratios and direction cosines is essential in three-dimensional geometry, as it is used in solving many problems related to the direction and orientation of lines and planes in 3D space.

**Describe Direction Ratios and Direction Cosines of a Line**

In three-dimensional geometry, the direction ratios and direction cosines of a line are two important concepts used to determine the direction of a line. Direction ratios are used to determine the direction of a line by expressing the ratios of the differences of the coordinates of two points on the line in a suitable form. Direction cosines are used to express the direction of a line in terms of the angles it makes with the positive x, y, and z-axes. In this topic, we will learn about the direction ratios and direction cosines of a line.

Direction Ratios of a Line:

The direction ratios of a line are the ratios of the differences of the coordinates of two points on the line in a suitable form. If (x1, y1, z1) and (x2, y2, z2) are two points on a line, then the direction ratios of the line can be calculated using the following formula:

Direction ratio of x = (x_{2} – x_{1})

Direction ratio of y = (y_{2} – y_{1})

Direction ratio of z = (z_{2 }– z_{1})

It is important to note that the direction ratios of a line are not unique, as they can be multiplied by a non-zero constant to give different sets of direction ratios that still represent the same line.

Direction Cosines of a Line:

The direction cosines of a line are the cosines of the angles it makes with the positive x, y, and z-axes. If (l, m, n) are the direction cosines of a line, then they can be calculated using the following formula:

l = cos(θ x)

m = cos(θ y)

n = cos(θ z)

where θ x, θ y, and θ z are the angles that the line makes with the positive x, y, and z-axes, respectively.

It is important to note that the direction cosines of a line are unique and are independent of the choice of the points on the line.

Relationship between Direction Ratios and Direction Cosines:

The direction cosines and direction ratios of a line are related by the following formula:

l^{2} + m^{2} + n^{2 }= 1

This formula can be derived by using the fact that the cosine squared of the angle a line makes with a given axis is equal to the square of the corresponding direction ratio divided by the sum of the squares of all three direction ratios.

Conclusion:

The direction ratios and direction cosines of a line are two important concepts used to determine the direction of a line in three-dimensional geometry. The direction ratios can be calculated using the differences of the coordinates of two points on the line, while the direction cosines can be calculated using the angles that the line makes with the positive x, y, and z-axes. The direction cosines of a line are unique and are independent of the choice of the points on the line, while the direction ratios are not unique and can be multiplied by a non-zero constant to give different sets of direction ratios that still represent the same line. The relationship between direction ratios and direction cosines is given by a formula that relates the sum of the squares of the direction cosines to 1.

**Describe the Equation of a Line through a given point and parallel to the given Vector**

In three-dimensional geometry, the equation of a line is a fundamental concept used to represent a line in three-dimensional space. The equation of a line through a given point and parallel to a given vector is a specific type of equation that is useful for solving problems involving lines in three-dimensional space. In this topic, we will learn how to describe the equation of a line through a given point and parallel to a given vector.

Equation of a Line through a Given Point and Parallel to a Given Vector:

Let (x_{1}, y_{1}, z_{1}) be a given point and let a = (a1, a2, a3) be a given vector parallel to the line. Then the equation of the line passing through the point (x1, y1, z1) and parallel to the vector a can be expressed in vector form as:

r = (x_{1}, y_{1}, z_{1}) + t * a

where r = (x, y, z) is any point on the line, and t is a parameter.

The above equation is known as the vector equation of a line.

To express the equation of the line in Cartesian form, we need to eliminate the parameter t. To do this, we can use the following system of equations:

x = x_{1} + a_{1}t

y = y_{1} + a_{2}t

z = z_{1} + a_{3}t

Solving for t in any one of these equations, we get:

t = (x – x_{1})/a_{1} = (y – y_{1})/a_{2} = (z – z_{1})/a_{3}

Substituting this value of t in the other two equations, we get:

(x – x_{1})/a_{1} = (y – y_{1})/a_{2} = (z – z_{1})/a_{3}

This is known as the Cartesian equation of the line.

Example:

Find the equation of the line passing through the point (1, 2, 3) and parallel to the vector a = (2, -1, 4).

Solution:

Using the vector equation of a line, we have:

r = (x_{1}, y_{1}, z_{1}) + t * a

r = (1, 2, 3) + t * (2, -1, 4)

To express the equation of the line in Cartesian form, we use the system of equations:

x = 1 + 2t

y = 2 – t

z = 3 + 4t

Solving for t in any one of these equations, we get:

t = (x – 1)/2 = (y – 2)/(-1) = (z – 3)/4

Substituting this value of t in the other two equations, we get:

x – 1 = 2t

y – 2 = -t

z – 3 = 4t

Multiplying the second equation by 2 and adding it to the first equation, we get:

3y – 4 = 0

Multiplying the third equation by 2 and adding it to the first equation, we get:

7z – 10 = 0

So the Cartesian equation of the line is:

(x – 1)/2 = (y – 2)/(-1) = (z – 3)/4 = t

3y – 4 = 0

7z – 10 = 0

Conclusion:

The equation of a line through a given point and parallel to a given vector is an important concept in three-dimensional geometry

**Describe the Equation of a Line passing through two given Points**

In three-dimensional geometry, a line passing through two given points is an essential concept that is frequently used in various applications such as computer graphics, engineering, and physics. In this topic, we will learn how to describe the equation of a line passing through two given points.

Equation of a Line passing through two given Points:

Let P1 = (x_{1}, y_{1}, z_{1}) and P2 = (x_{2}, y_{2}, z_{2}) be two given points. Then the equation of the line passing through P1 and P2 can be expressed in vector form as:

r = P_{1 }+ t * (P_{2} – P_{1})

where r = (x, y, z) is any point on the line, and t is a parameter.

The above equation is known as the vector equation of a line passing through two given points.

To express the equation of the line in Cartesian form, we need to eliminate the parameter t. To do this, we can use the following system of equations:

x = x_{1} + (x_{2} – x_{1})t

y = y_{1} + (y_{2} – y_{1})t

z = z_{1} + (z_{2 }– z_{1)}t

Solving for t in any one of these equations, we get:

t = (x – x1)/(x2 – x1) = (y – y1)/(y2 – y1) = (z – z1)/(z2 – z1)

Substituting this value of t in the other two equations, we get:

(x – x1)/(x2 – x1) = (y – y1)/(y2 – y1) = (z – z1)/(z2 – z1)

This is known as the Cartesian equation of the line passing through two given points.

Example:

Find the equation of the line passing through the points P1 = (1, 2, 3) and P2 = (4, 5, 6).

Solution:

Using the vector equation of a line passing through two given points, we have:

r = P1 + t * (P2 – P1)

r = (1, 2, 3) + t * (4, 5, 6) – (1, 2, 3)

r = (1, 2, 3) + t * (3, 3, 3)

To express the equation of the line in Cartesian form, we use the system of equations:

x = 1 + 3t

y = 2 + 3t

z = 3 + 3t

Solving for t in any one of these equations, we get:

t = (x – 1)/3 = (y – 2)/3 = (z – 3)/3

Substituting this value of t in the other two equations, we get:

(x – 1)/3 = (y – 2)/3 = (z – 3)/3

So the Cartesian equation of the line passing through the points P1 and P2 is:

(x – 1)/3 = (y – 2)/3 = (z – 3)/3

Conclusion:

The equation of a line passing through two given points is a fundamental concept in three-dimensional geometry that has numerous applications.

**Recall the Angle between two Lines**

In three-dimensional geometry, the angle between two lines is an essential concept that is frequently used in various applications such as computer graphics, engineering, and physics. In this topic, we will learn how to recall the angle between two lines.

Angle between two Lines:

Let L1 and L2 be two non-vertical lines in three-dimensional space. Let θ be the angle between them. We can find the angle θ between the two lines using the following formula:

cos θ = (L1 · L2) / (|L1| * |L2|)

where L1 · L2 is the dot product of the direction vectors of the lines L1 and L2, and |L1| and |L2| are the magnitudes of the direction vectors.

If the lines are parallel, then the angle between them is 0° or 180°. If the lines intersect, then the angle between them is acute, and if they are skew (i.e., they are neither parallel nor intersecting), then the angle between them is obtuse.

Example:

Find the angle between the lines L1 and L2, where L1 passes through the points (1, 2, 3) and (4, 5, 6), and L2 passes through the points (0, 0, 1) and (1, 0, 0).

Solution:

To find the angle between the lines L1 and L2, we need to find the direction vectors of these lines.

The direction vector of L1 is:

L1 = (4, 5, 6) – (1, 2, 3) = (3, 3, 3)

The direction vector of L2 is:

L2 = (1, 0, 0) – (0, 0, 1) = (1, 0, -1)

The dot product of L1 and L2 is:

L1 · L2 = (3 * 1) + (3 * 0) + (3 * -1) = 0

The magnitude of L1 is:

|L1| = sqrt(3^{2} + 3^{2} + 3^{2}) = 3sqrt(3)

The magnitude of L2 is:

|L2| = sqrt(1^{2} + 0^{2 }+ (-1)^{2}) = sqrt(2)

Substituting these values in the formula, we get:

cos θ = (L1 · L2) / (|L1| * |L2|) = 0 / (3sqrt(3) * sqrt(2)) = 0

Therefore, the angle between the lines L1 and L2 is 90°.

Conclusion:

The angle between two lines is a fundamental concept in three-dimensional geometry that has numerous applications. It helps us to determine whether two lines are parallel, intersecting or skew, and also enables us to calculate the angle between them.

**Recall the distance between two Skew Lines and two Parallel Lines**

In three-dimensional geometry, the distance between two lines is an essential concept that is frequently used in various applications such as computer graphics, engineering, and physics. In this topic, we will learn how to recall the distance between two skew lines and two parallel lines.

Distance between two Skew Lines:

Two non-intersecting lines in 3D space are called skew lines. The distance between two skew lines is the shortest distance between any point on one line to the other line. The formula to calculate the distance between two skew lines is given by:

d = |(P2 – P1) · n|

where P1 and P2 are two arbitrary points on the two lines, and n is a normal vector to both lines. The distance d is the perpendicular distance between the two skew lines.

Example:

Find the distance between the skew lines L1 and L2, where L1 passes through the points (1, 2, 3) and (4, 5, 6), and L2 passes through the points (0, 0, 1) and (1, 0, 0).

Solution:

To find the distance between the skew lines L1 and L2, we need to find a normal vector to both lines. We can find a normal vector by taking the cross product of the direction vectors of the two lines.

The direction vector of L1 is:

L1 = (4, 5, 6) – (1, 2, 3) = (3, 3, 3)

The direction vector of L2 is:

L2 = (1, 0, 0) – (0, 0, 1) = (1, 0, -1)

Taking the cross product of L1 and L2, we get:

n = L1 x L2 = (3, 3, 3) x (1, 0, -1) = (3, 6, -3)

Now, we need to find any point on each line. Let’s take (1, 2, 3) on L1 and (0, 0, 1) on L2.

Substituting these values in the formula, we get:

d = |(P2 – P1) · n| = |(0 – 1, 0 – 2, 1 – 3) · (3, 6, -3)| = |-2 · 3| = 6

Therefore, the distance between the skew lines L1 and L2 is 6 units.

Distance between two Parallel Lines:

Two lines in 3D space are called parallel lines if they have the same direction vectors or are coincident. The distance between two parallel lines is the shortest distance between any point on one line to the other line. The formula to calculate the distance between two parallel lines is given by:

d = |(P2 – P1) · n|

where P1 and P2 are two arbitrary points on the two lines, and n is a normal vector to both lines. The distance d is the perpendicular distance between the two parallel lines.

**Describe the Angle between two Planes**

Description: The angle between two planes is the angle formed by their normal vectors. In this learning outcome, we will describe how to find the angle between two planes and provide suitable examples.

- Finding the Normal Vector of a Plane:

Before we can calculate the angle between two planes, we need to find their normal vectors. The normal vector of a plane is a vector perpendicular to the plane’s surface. We can find the normal vector of a plane in several ways, including using the cross-product of two vectors in the plane or using the coefficients of the plane’s equation.

- Calculating the Angle between Two Planes:

Once we have found the normal vectors of the two planes, we can calculate the angle between them using the dot product formula:

θ = cos⁻¹[(n1 . n2)/(|n1| |n2|)]

where n1 and n2 are the normal vectors of the two planes, and |n1| and |n2| are the magnitudes of the normal vectors.

Examples:

Find the angle between the planes 2x – y + z = 0 and x + 2y – 3z = 0.

Solution:

The normal vector of the first plane is (2, -1, 1), and the normal vector of the second plane is (1, 2, -3). Using the dot product formula, we get:

θ = cos⁻¹[(2*-1 + -1*2 + 1*-3)/(√(2²+(-1)²+1²) √(1²+2²+(-3)²))]

= cos⁻¹[-7/√6 * √14]

= 132.1 degrees (approx.)

Example

Find the angle between the planes x + 2y + z = 3 and x + y – z = 0.

Solution:

The normal vector of the first plane is (1, 2, 1), and the normal vector of the second plane is (1, 1, -1). Using the dot product formula, we get:

θ = cos⁻¹[(1*1 + 2*1 + 1*-1)/(√(1²+2²+1²) √(1²+1²+(-1)²))]

= cos⁻¹[2/√6 * √3]

= 24.9 degrees (approx.)

In summary, the angle between two planes can be found by first calculating their normal vectors and then using the dot product formula. The angle between two planes is an important concept in mathematics and is used in various fields, such as computer graphics and engineering.

**Describe the Angle between a Line and a Plane**

Description: In this learning outcome, we will describe how to find the angle between a line and a plane and provide suitable examples.

- Finding the Normal Vector of a Plane:

Before we can calculate the angle between a line and a plane, we need to find the normal vector of the plane. The normal vector of a plane is a vector perpendicular to the plane’s surface. We can find the normal vector of a plane in several ways, including using the cross-product of two vectors in the plane or using the coefficients of the plane’s equation.

- Finding the Direction Vector of a Line:

The direction vector of a line is a vector that points in the direction of the line. We can find the direction vector of a line by taking the difference between two points on the line.

- Calculating the Angle between a Line and a Plane:

Once we have found the normal vector of the plane and the direction vector of the line, we can calculate the angle between them using the dot product formula:

θ = cos⁻¹[(n . d)/(|n| |d|)]

where n is the normal vector of the plane, d is the direction vector of the line, and |n| and |d| are the magnitudes of the normal vector and the direction vector, respectively.

Examples:

Find the angle between the line with equation x = 1 + t, y = 2 – t, z = 3 + 2t and the plane with equation 2x – y + z = 4.

Solution:

The normal vector of the plane is (2, -1, 1), and the direction vector of the line is (1, -1, 2). Using the dot product formula, we get:

θ = cos⁻¹[(2*1 + -1*-1 + 1*2)/(√(2²+(-1)²+1²) √(1²+(-1)²+2²))]

= cos⁻¹[3/√6 * √6]

= 45 degrees (approx.)

Example

Find the angle between the line passing through the points (1, 2, 3) and (2, 3, 5) and the plane with equation x – y + 2z = 1.

Solution:

The normal vector of the plane is (1, -1, 2), and the direction vector of the line is (1, 1, 2). Using the dot product formula, we get:

θ = cos⁻¹[(1*1 + -1*1 + 2*2)/(√(1²+(-1)²+2²) √(1²+1²+2²))]

= cos⁻¹[4/√6 * √6]

= 45 degrees (approx.)

In summary, the angle between a line and a plane can be found by first calculating the normal vector of the plane and the direction vector of the line and then using the dot product formula. The angle between a line and a plane is an important concept in mathematics and is used in various fields, such as computer graphics and engineering.

**Describe the Coplanarity of two lines**

- Coplanarity:

Two lines are said to be coplanar if they lie in the same plane. This means that the lines do not intersect and do not diverge away from each other. Instead, they remain in the same plane and may be parallel or intersect at a point.

- Vector Equation of a Line:

A line can be represented by a vector equation in the form of r = a + tb, where a is a point on the line, b is a vector in the direction of the line, and t is a parameter.

- Determining Coplanarity of Two Lines:

To determine if two lines are coplanar, we need to check if the direction vectors of the lines are parallel or if they intersect at a point. If the direction vectors are parallel, the lines are coplanar. If the direction vectors intersect at a point, the lines may or may not be coplanar. To determine if they are coplanar, we need to check if a third point lies on the same plane as the two points of intersection.

Example1:

Determine if the lines with vector equations r1 = (1, 2, 3) + t(2, 1, 1) and r2 = (3, 5, 7) + t(4, 2, 2) are coplanar.

Solution:

The direction vector of the first line is (2, 1, 1), and the direction vector of the second line is (4, 2, 2). Since these direction vectors are parallel, the lines are coplanar.

Example2:

Determine if the lines with vector equations r1 = (1, 2, 3) + t(2, 1, 1) and r2 = (3, 5, 7) + t(1, 1, 1) are coplanar.

Solution:

The direction vector of the first line is (2, 1, 1), and the direction vector of the second line is (1, 1, 1). These direction vectors do not appear to be parallel, so we need to find their point of intersection. Setting the two vector equations equal to each other, we get:

(1, 2, 3) + t(2, 1, 1) = (3, 5, 7) + s(1, 1, 1)

Solving for t and s, we get t = 1 and s = 2. Substituting these values into either equation, we get the point of intersection as (3, 4, 5). To determine if the lines are coplanar, we need to check if another point lies on the same plane as the two points of intersection. Let’s take a third point on the first line as (3, 4, 5) + r(2, 1, 1). Substituting this into the equation of the second line, we get:

(3, 4, 5) + r(2, 1, 1) = (3, 5, 7) + s(1, 1, 1)

Solving for r and s, we get r = 1 and s = 2. Since we get the same values of t and s, the lines are coplanar.

**Recall the distance formulae of a Point from a Plane**

- Distance between a Point and a Plane:

In 3D space, the distance between a point and a plane can be found using the formula:

d = |ax + by + cz + d| / sqrt(a^{2} + b^{2} + c^{2})

where (x, y, z) is the coordinates of the point, a, b, and c are the coefficients of the equation of the plane, and d is the constant term in the equation of the plane.

- Proof of the Formula:

Let P be the point (x, y, z), and let Q be the point on the plane that is closest to P. Then, the vector PQ is perpendicular to the plane. We can find the equation of this vector using the point-to-plane formula:

ax + by + cz + d = 0

This is the equation of the plane. If we substitute the coordinates of point Q into this equation, we get:

a(x_{Q}) + b(y_{Q}) + c(z_{Q}) + d = 0

Solving for Q, we get:

Q = (x_{Q}, y_{Q}, z_{Q}) = (x – ad/sqrt(a^{2 }+ b^{2 }+ c^{2}), y – bd/sqrt(a^{2} + b^{2} + c^{2}), z – cd/sqrt(a^{2} + b^{2} + c^{2}))

The distance between P and Q is the length of vector PQ, which can be found using the formula:

|PQ| = |(x – x_{Q}, y – y_{Q}, z – z_{Q})| = |ax + by + cz + d| / sqrt(a2 + b^{2} + c^{2})

Example 1: Find the distance between the point (1, 2, 3) and the plane with equation 2x + 3y – 4z + 5 = 0.

Solution:

The coefficients of the equation of the plane are a = 2, b = 3, and c = -4, and the constant term is d = -5. Substituting these values into the formula, we get:

d = |2(1) + 3(2) – 4(3) – 5| / sqrt(2^{2} + 3^{2} + (-4)^{2}) = 6 / sqrt(29)

Example 2: Find the distance between the point (2, -1, 4) and the plane with equation x + 2y – z + 6 = 0.

Solution:

The coefficients of the equation of the plane are a = 1, b = 2, and c = -1, and the constant term is d = -6. Substituting these values into the formula, we get:

d = |1(2) + 2(-1) – 1(4) – 6| / sqrt(1^{2} + 2^{2} + (-1)^{2}) = 5 / sqrt(6)

In summary, the distance between a point and a plane in 3D space can be found using the formula: d = |ax + by + cz + d| / sqrt(a^{2} + b^{2} + c^{2}).

**Describe the equations of a Plane: In Normal Form, Perpendicular to a given Vector and passing through a Point, and In Intercept Form**

A plane is a two-dimensional flat surface that extends infinitely in all directions. It can be described by several different equations depending on the information available about the plane. In this learning outcome, we will discuss the equations of a plane in normal form, perpendicular to a given vector and passing through a point, and in intercept form.

- Normal form:

A plane in normal form is described by the equation:

Ax + By + Cz + D = 0

where A, B, and C are the coefficients of the variables x, y, and z respectively, and D is a constant. The vector (A, B, C) is called the normal vector of the plane, and it is perpendicular to every vector that lies in the plane.

Example: Find the equation of the plane that passes through the points (1, 2, 3), (4, 5, 6), and (7, 8, 9).

To find the normal vector of the plane, we take the cross product of the vectors between any two points on the plane. For example, we can take the vectors (1, 2, 3) to (4, 5, 6) and (1, 2, 3) to (7, 8, 9):

(4-1, 5-2, 6-3) = (3, 3, 3)

(7-1, 8-2, 9-3) = (6, 6, 6)

The cross product of these vectors is:

(3, 3, 3) x (6, 6, 6) = (0, 18, -18)

So the normal vector of the plane is (0, 18, -18). To find the equation of the plane, we substitute this normal vector and one of the points into the normal form equation:

0x + 18y – 18z + D = 0

Substituting (1, 2, 3) gives:

0 + 18(2) – 18(3) + D = 0

D = -12

So the equation of the plane in normal form is:

0x + 18y – 18z – 12 = 0

- Perpendicular to a given vector and passing through a point:

A plane can also be described by an equation in which the normal vector is given by a known vector and a point that the plane passes through. The equation of a plane in this form is:

(A – x) * nx + (B – y) * ny + (C – z) * nz = 0

where (A, B, C) is a point that the plane passes through, and (nx, ny, nz) is a vector perpendicular to the plane.

Example: Find the equation of the plane that passes through the point (2, 1, -1) and is perpendicular to the vector (1, 2, 3).

We use the point (2, 1, -1) as the point the plane passes through, and the vector (1, 2, 3) as the normal vector. Substituting these values into the equation above, we get:

(1 – 2) * 1 + (2 – 1) * 2 + (3 – (-1)) * 3 = 0

Simplifying, we get:

-1 + 2 + 9 = 0

So the equation of the plane is:

-x + 2y + 3z – 3 = 0

**Describe the equation of a Circle**

A circle is a two-dimensional geometric shape that consists of all the points that are equidistant from a given point called the center. The distance from the center to any point on the circle is called the radius. There are different ways to describe the equation of a circle, but the most common form is the standard form, which is:

(x – a)^{2} + (y – b)^{2} = r^{2}

where (a, b) is the center of the circle, and r is the radius.

Example: Find the equation of the circle with center (2, 3) and radius 4.

We can directly substitute the values of the center and radius into the standard form equation to obtain:

(x – 2)^{2} + (y – 3)^{2} = 4^{2}

Expanding the terms, we get:

x^{2} – 4x + 4 + y^{2} – 6y + 9 = 16

Simplifying, we get:

x^{2} + y^{2} – 4x – 6y – 3 = 0

This is the equation of the circle in standard form.

Another way to describe the equation of a circle is in the general form, which is:

x^{2} + y^{2} + Dx + Ey + F = 0

where D, E, and F are constants that depend on the center and radius of the circle.

Example: Find the equation of the circle with center (3, -2) and radius 5.

We can start by substituting the values of the center and radius into the standard form equation and expanding the terms:

(x – 3)^{2} + (y + 2)^{2 }= 5^{2}

x^{2} – 6x + 9 + y^{2} + 4y + 4 = 25

x^{2} + y^{2 }– 6x + 4y – 12 = 0

Now we can rearrange this equation to match the general form by completing the square for x and y:

(x^{2} – 6x + 9) + (y^{2} + 4y + 4) – 25 = 0

(x – 3)^{2} + (y + 2)^{2} – 25 = 0

x^{2} – 6x + 9 + y^{2} + 4y + 4 – 25 = 0

x^{2} + y^{2} – 6x + 4y – 12 = 0

Comparing this with the general form equation, we see that D = -6, E = 4, and F = -12. Therefore, the equation of the circle in general form is:

x^{2 }+ y^{2} – 6x + 4y – 12 = 0.

**Find the center and radius of a given Circle**

Given the equation of a circle in the standard form, (x-a)^{2} + (y-b)^{2} = r^{2}, we can find the center and radius of the circle.

Center: The center of the circle is represented by the point (a, b).

Radius: The radius of the circle is represented by the value r.

Example: Find the center and radius of the circle given by the equation (x-3)^{2 }+ (y+4)^{2 }= 16.

We can compare the given equation with the standard form equation:

(x-a)^{2} + (y-b)^{2} = r^{2}

We can see that a = 3, b = -4, and r = 4.

Therefore, the center of the circle is (3, -4), and the radius is 4 units.

Another way to find the center and radius of a circle is to convert the equation of the circle from standard form to general form.

In the general form of the equation of a circle, x^{2} + y^{2} + Dx + Ey + F = 0, the center of the circle is represented by the point (-D/2, -E/2), and the radius is represented by the value sqrt(D^{2} + E^{2} – 4F)/2.

Example: Find the center and radius of the circle given by the equation x^{2} + y^{2} – 4x + 6y + 12 = 0.

We can convert this equation from general form to standard form by completing the square for x and y:

x^{2} – 4x + 4 + y^{2} + 6y + 9 – 12 = 0

(x-2)^{2} + (y+3)^{2} = 1

Now we can see that the center of the circle is (2, -3) and the radius is 1 unit.

Therefore, we can find the center and radius of a given circle either by comparing the given equation with the standard form equation or by converting the given equation from general form to standard form.

**Describe the diametric form of a Circle**

The diametric form of a circle is a different way of representing a circle in which two opposite endpoints of the diameter are used to define the equation of the circle. In other words, the diametric form represents the circle in terms of its diameter.

Let P and Q be two opposite points on a circle with diameter PQ, and let the midpoint of PQ be M. Then, the equation of the circle in diametric form is:

(x – Mx)(x – Px) + (y – My)(y – Py) = 0

where Mx and My are the coordinates of the midpoint M, and Px and Py are the coordinates of one of the endpoints, say P.

Alternatively, we can use Q instead of P to obtain the following equation:

(x – Mx)(x – Qx) + (y – My)(y – Qy) = 0

Example: Find the equation of the circle with endpoints (2, 4) and (6, -2).

First, we find the midpoint of the line segment joining the two points, which is (4, 1).

Next, we use one of the endpoints, say (2, 4), to obtain the equation in diametric form:

(x – 4)(x – 2) + (y – 1)(y – 4) = 0

Simplifying this equation, we get:

x^{2} – 6x + y^{2} – 5y + 14 = 0

Therefore, the equation of the circle with endpoints (2, 4) and (6, -2) in diametric form is x^{2} – 6x + y^{2} – 5y + 14 = 0.

The advantage of using the diametric form of a circle is that it can be used to find the equation of a circle even if its center is not given. This form is also useful in solving problems related to chords and tangents of a circle.

**Find the equation of a Circle passing through three non-collinear points**

To find the equation of a circle passing through three non-collinear points, we can use the following steps:

- Find the perpendicular bisectors of any two of the three given points. The perpendicular bisector is a line that passes through the midpoint of the line segment joining two points and is perpendicular to that line segment.
- The intersection of the two perpendicular bisectors is the center of the circle.
- Calculate the radius of the circle by finding the distance between the center and one of the given points.
- Write the equation of the circle using the center and radius.

Example: Find the equation of the circle passing through the points (1, 2), (3, 4), and (5, 6).

Step 1: Find the perpendicular bisectors of two of the given points.

We can choose any two points to find the perpendicular bisectors. Let’s choose (1, 2) and (3, 4).

The midpoint of the line segment joining (1, 2) and (3, 4) is ((1+3)/2, (2+4)/2) = (2, 3).

The slope of the line segment is (4-2)/(3-1) = 1, so the slope of the perpendicular bisector is -1.

The equation of the perpendicular bisector is y – 3 = -1(x – 2) or x + y = 5.

Similarly, the perpendicular bisector of the line segment joining (3, 4) and (5, 6) is x – y = -1.

Step 2: Find the intersection of the perpendicular bisectors.

Solving the system of equations x + y = 5 and x – y = -1, we get x = 2 and y = 3. Therefore, the center of the circle is (2, 3).

Step 3: Calculate the radius of the circle.

The radius of the circle is the distance between the center and one of the given points. Let’s choose (1, 2).

The distance between (1, 2) and (2, 3) is sqrt((2-1)^{2} + (3-2)^{2}) = sqrt(2).

Step 4: Write the equation of the circle.

The equation of the circle is (x-2)^{2 }+ (y-3)^{2} = (sqrt(2))^{2}, which simplifies to (x-2)^{2} + (y-3)^{2} = 2.

Therefore, the equation of the circle passing through the points (1, 2), (3, 4), and (5, 6) is (x-2)^{2} + (y-3)^{2} = 2.

**Describe the intercepts made on the axes by a Circle**

The intercepts made by a circle on the x-axis and y-axis are known as the x-intercept and y-intercept, respectively. The x-intercept is the point at which the circle intersects the x-axis, while the y-intercept is the point at which the circle intersects the y-axis. These intercepts can be found using the equation of the circle.

If the equation of the circle is given in standard form, (x-a)^{2} + (y-b)^{2} = r^{2}, then the x-intercept and y-intercept can be found by setting y = 0 and x = 0, respectively.

Setting y = 0, we get (x-a)^{2} + (0-b)^{2} = r^{2}, which simplifies to (x-a)^{2} = r^{2} – b^{2}. Taking the square root of both sides, we get x-a = ±sqrt(r^{2}-b^{2}).^{ }Therefore, the x-intercepts are (a+sqrt(r^{2}-b2), 0) and (a-sqrt(r^{2}-b^{2}), 0).

Similarly, setting x = 0, we get (0-a)^{2} + (y-b)^{2} = r^{2}, which simplifies to (y-b)^{2} = r^{2} – a^{2}. Taking the square root of both sides, we get y-b = ±sqrt(r^{2}-a^{2}). Therefore, the y-intercepts are (0, b+sqrt(r^{2}-a^{2})) and (0, b-sqrt(r^{2}-a^{2})).

Example: Find the x-intercepts and y-intercepts of the circle with center (2, 3) and radius 4.

The equation of the circle in standard form is (x-2)^{2} + (y-3)^{2} = 4^{2}.

Setting y = 0, we get (x-2^{2} + (0-3)^{2} = 4^{2}, which simplifies to (x-2)^{2} = 7^{2}. Taking the square root of both sides, we get x-2 = ±7. Therefore, the x-intercepts are (9, 0) and (-5, 0).

Setting x = 0, we get (0-2)^{2} + (y-3)^{2} = 4^{2}, which simplifies to (y-3)^{2} = 7^{2}. Taking the square root of both sides, we get y-3 = ±7. Therefore, the y-intercepts are (0, 10) and (0, -4).

Therefore, the x-intercepts are (9, 0) and (-5, 0), and the y-intercepts are (0, 10) and (0, -4).

**Find the equation of a Circle when touches or cut the axes**

A circle that touches or cuts the axes has one or both intercepts equal to the radius of the circle. The equation of such a circle can be found by using the intercepts or the point of contact.

Case 1: Circle touches one axis

If a circle touches one axis, say the x-axis, at point (r,0), then the equation of the circle can be written as (x-r)^{2} + y^{2} = r^{2}, where r is the radius of the circle.

Similarly, if a circle touches the y-axis at point (0,r), then the equation of the circle can be written as x^{2} + (y-r)^{2} = r^{2}.

Case 2: Circle cuts both axes

If a circle cuts both the x and y-axes, then it passes through the origin (0,0). Let r be the radius of the circle, and let (a,0) and (0,b) be the x and y intercepts, respectively. Then, we have:

a^{2} + b^{2} = r^{2} (equation 1)

The center of the circle lies on the line joining the midpoints of the x and y intercepts, which is the line y = x. Therefore, the center of the circle is (h, h) or (-h, -h), where h is the distance from the origin to the center of the circle.

Using the distance formula, we have:

h^{2} + h^{2} = r^{2}

2h^{2} = r^{2}

h = ±(r/√2)

If the center of the circle is (h,h), then the equation of the circle is (x-h)^{2 }+ (y-h)^{2} = r^{2}.

If the center of the circle is (-h,-h), then the equation of the circle is (x+h)^{2} + (y+h)^{2} = r^{2}.

Example: Find the equation of the circle which touches the x-axis at (3,0) and cuts the y-axis at (0,4).

Let r be the radius of the circle, and let the center of the circle be (h,h).

The x-intercept is (3,0), so (3-h)^{2} + h^{2} = r^{2}. Simplifying, we get 10h – 9r = 0.

The y-intercept is (0,4), so h^{2} + (4-h)^{2} = r^{2}. Simplifying, we get 2h^{2} – 8h + 16 = r^{2}.

Using equation 1, we get a^{2} + b^{2} = r^{2}, where a = 3 and b = 4. Therefore, 9 + 16 = r2, so r = 5.

Substituting r = 5 in the first equation, we get 10h – 45 = 0, so h = 4.5.

Therefore, the center of the circle is (4.5, 4.5), and the equation of the circle is (x-4.5)^{2} + (y-4.5)^{2 }= 5^{2}.

**Describe the concept of Conic Section**

Conic sections are a set of curves that can be obtained by intersecting a cone with a plane. The four main types of conic sections are the circle, ellipse, parabola, and hyperbola.

Each type has unique properties and characteristics.

- Circle: A circle is a conic section formed by intersecting a cone with a plane that is perpendicular to the axis of the cone. It is a closed curve consisting of all points equidistant from a fixed center point. The distance from the center to any point on the circle is called the radius.
- Ellipse: An ellipse is a conic section formed by intersecting a cone with a plane that is slanted with respect to the axis of the cone. It is a closed curve consisting of all points such that the sum of the distances from two fixed points, called the foci, to any point on the curve is constant. The major axis is the longest diameter of the ellipse, and the minor axis is the shortest diameter.
- Parabola: A parabola is a conic section formed by intersecting a cone with a plane that is parallel to one of the generating lines of the cone. It is a curved shape with a reflective property: all light rays parallel to the axis of symmetry of the parabola are reflected off the surface and converge at a single point called the focus. The vertex is the point where the parabola makes its sharpest turn.
- Hyperbola: A hyperbola is a conic section formed by intersecting a cone with a plane that is at an angle to the axis of the cone. It is a curve consisting of two distinct branches, each with its own focus and directrix. The distance between the foci divided by the distance between the vertices is a constant called the eccentricity.

Conic sections have applications in various fields, including mathematics, physics, engineering, and astronomy. They exhibit unique properties and mathematical relationships that make them valuable in understanding geometric shapes, motion, optics, and celestial bodies.

**Recall the term Eccentricity**

Eccentricity is a term used to describe how “squished” or elongated a conic section is. It is a measure of the distance between the center of the conic section and one of its foci, relative to the length of the major axis. The eccentricity is denoted by the symbol e and is defined as:

e = c/a

where c is the distance from the center of the conic section to one of its foci, and a is the length of the semi-major axis. For an ellipse, the eccentricity is a number between 0 and 1, and it describes how much the ellipse is “squished” or elongated. When e = 0, the ellipse is a circle, and when e = 1, the ellipse is a line segment.

For a hyperbola, the eccentricity is a number greater than 1, and it describes how much the hyperbola is elongated. When e = 1, the hyperbola is a line segment, and when e > 1, the hyperbola has two branches that are “very far apart.”

For a parabola, the eccentricity is exactly 1, and it describes how “flat” or “shallow” the parabola is. A parabola with a small eccentricity is very narrow and steep, while a parabola with a large eccentricity is very flat and wide.

Eccentricity is an important concept in the study of conic sections because it allows us to describe and compare different types of conic sections. It is also used in many practical applications, such as designing satellite orbits and reflecting telescopes.

**Recall the nature of Conics in view of Second Degree Equation**

The second-degree equation in two variables, also known as a quadratic equation, has the general form:

Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0

where A, B, C, D, E, and F are constants, and x and y are variables. When graphed in the xy-plane, this equation represents a conic section. The nature of the conic section depends on the values of the coefficients A, B, and C.

- If A and C have the same sign (both positive or both negative), the conic section is an ellipse. The center of the ellipse is (-D/2A, -E/2C), and the major and minor axes are given by:

a = sqrt[(D^{2} – 4AC)/4A^{2} + (E^{2} – 4AC)/4C^{2}]

b = sqrt[(D^{2} – 4AC)/4A^{2} – (E^{2} – 4AC)/4C^{2}]

The eccentricity e of the ellipse is given by:

e = sqrt[(a^{2} – b^{2})/a^{2}]

- If A and C have opposite signs, the conic section is a hyperbola. The center of the hyperbola is (-D/2A, -E/2C), and the transverse and conjugate axes are given by:

a = sqrt[(4AC – D^{2})/4A^{2} – (E^{2} – 4AC)/4C2]

b = sqrt[(4AC – D^{2})/4A^{2} + (E^{2} – 4AC)/4C^{2}]

The eccentricity e of the hyperbola is given by:

e = sqrt[(a^{2} + b2)/a^{2}]

- If A and C are both zero and B is nonzero, the conic section is a parabola. The axis of the parabola is either vertical or horizontal, depending on the sign of B. The vertex of the parabola is given by (-D/2B, -E/2B), and the equation of the directrix is either x = D/2A – 1/2B or y = E/2C – 1/2B, depending on the orientation of the axis.
- If A = C = 0 and B = 0, the equation reduces to a linear equation, which represents a straight line. The slope and y-intercept of the line can be found by solving for y in terms of x.

In summary, the nature of a conic section depends on the coefficients of its second-degree equation, with A and C determining whether it is an ellipse or hyperbola, and B determining whether it is a parabola or a straight line.

**Find the Centre of Conics**

The center of a conic section is a point that lies at the intersection of the major and minor axes. The location of the center depends on the type of conic section and the coefficients of its second-degree equation.

- Ellipse: The center of an ellipse is the midpoint of the line segment joining the two foci. If the equation of the ellipse is given in standard form, i.e., in the form (x – h)
^{2}/a^{2}+ (y – k)^{2}/b^{2}= 1 or (y – k)^{2}/a^{2}+ (x – h)^{2}/b^{2}= 1, then the center is at (h, k).

For example, consider the ellipse given by the equation (x – 2)^{2}/16 + (y + 3)^{2}/9 = 1. The center of the ellipse is (2, -3).

- Hyperbola: The center of a hyperbola is the midpoint of the line segment joining the two foci. If the equation of the hyperbola is given in standard form, i.e., in the form (x – h)
^{2/}a^{2}– (y – k)^{2}/b^{2}= 1 or (y – k)^{2}/a^{2}– (x – h)^{2}/b^{2}= 1, then the center is at (h, k).

For example, consider the hyperbola given by the equation (x – 3)^{2}/9 – (y + 2)^{2}/4 = 1. The center of the hyperbola is (3, -2).

- Parabola: The center of a parabola is the vertex of the parabola. If the equation of the parabola is given in standard form, i.e., in the form y = a(x – h)
^{2}+ k or x = a(y – k)^{2}+ h, then the center is at (h, k).

For example, consider the parabola given by the equation y = 2(x – 1)^{2} + 3. The vertex of the parabola is (1, 3), which is also the center of the parabola.

In summary, the center of a conic section can be found by determining the midpoint of the line segment joining the two foci for ellipses and hyperbolas, and by finding the vertex for parabolas. If the equation of the conic section is given in standard form, the coordinates of the center can be read directly from the equation.

**Describe the concept of Parabola**

A parabola is a conic section that is defined as the set of all points in a plane that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. The axis of a parabola is a line that is perpendicular to the directrix and passes through the focus.

The shape of a parabola is a symmetrical U-shape, and it can be oriented either horizontally or vertically, depending on the orientation of its axis. The equation of a parabola is of the form y = ax^{2} + bx + c or x = ay^{2} + by + c, where a, b, and c are constants that determine the shape, position, and orientation of the parabola.

The vertex of a parabola is the point where the parabola intersects its axis, and it is also the minimum or maximum point of the parabola. The focus is located on the axis of the parabola and is equidistant from the vertex and the directrix. The directrix is a line that is equidistant from the vertex and the focus, and it is located on the opposite side of the vertex from the focus.

The standard form of the equation of a parabola with its vertex at the origin is y = 4px^{2}, where p is the distance from the vertex to the focus and to the directrix. The equation of a parabola with its vertex at (h, k) can be obtained by shifting the standard form equation by h units horizontally and k units vertically, resulting in the equation y = 4p(x – h)^{2} + k or x = 4p(y – k)^{2} + h.

The parabolic shape is found in many natural phenomena, including the trajectories of objects that are thrown or projected, the shapes of satellite dishes and reflectors, and the shape of the antennas used in radio and television broadcasting.

In summary, a parabola is a conic section that is defined by its focus, directrix, and axis, and it has a U-shape that is determined by the equation of the parabola. The vertex of a parabola is the minimum or maximum point, and it is located at the intersection of the parabola and its axis.

**Recall the basic terminologies of Parabola (Axis, Vertex, Focus, Directrix, Double Ordinate, Latus Rectum, Focal Chord, and Focal Distance)**

Parabolas have several terminologies associated with them. The following are the basic terminologies of parabola:

- Axis: The axis of a parabola is a line that passes through the vertex and is perpendicular to the directrix. It is the line of symmetry of the parabola.
- Vertex: The vertex is the point where the axis of the parabola intersects the parabola. It is the minimum or maximum point of the parabola.
- Focus: The focus is a fixed point on the axis of the parabola that is equidistant from the vertex and the directrix.
- Directrix: The directrix is a fixed line that is equidistant from the vertex and the focus, and is on the opposite side of the vertex from the focus.
- Double ordinate: A line that passes through the focus and is perpendicular to the axis is called a double ordinate. It intersects the parabola at two points that are equidistant from the focus.
- Latus rectum: The latus rectum is a line segment that passes through the focus and is perpendicular to the axis. It is the chord of the parabola that passes through the focus, and it is twice the focal distance.
- Focal chord: A focal chord is any chord of the parabola that passes through the focus.
- Focal distance: The focal distance is the distance from the vertex to the focus, or from the vertex to the directrix.

For example, consider the parabola y^{2} = 4x. The axis of the parabola is the x-axis, and the vertex is at the origin (0, 0). The focus is located at (1, 0), and the directrix is the line x = -1. The double ordinate is the line x = 1, and the latus rectum is the line y = 2. Any chord of the parabola that passes through the point (1, 0) is a focal chord, and the focal distance is 1.

In summary, the basic terminologies of a parabola include the axis, vertex, focus, directrix, double ordinate, latus rectum, focal chord, and focal distance. These terminologies help to define the shape and position of the parabola and are used in various applications in physics, engineering, and other fields.

**Describe all standard forms of a Parabola**

A parabola is a conic section that is defined as the locus of all points in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix. There are three standard forms of a parabola, which are as follows:

- Vertex form: The vertex form of a parabola is given by the equation y = a(x – h)² + k, where (h, k) is the vertex of the parabola and a is a constant that determines the shape of the parabola. If a is positive, the parabola opens upward, and if a is negative, the parabola opens downward.

For example, the graph of the equation y = 2(x – 3)² + 4 is a parabola that opens upward with vertex (3, 4) and axis of symmetry x = 3.

- Intercept form: The intercept form of a parabola is given by the equation y = a(x – p)(x – q), where p and q are the x-intercepts of the parabola and a is a constant that determines the shape of the parabola. If a is positive, the parabola opens upward, and if a is negative, the parabola opens downward.

For example, the graph of the equation y = -(x – 1)(x – 3) is a parabola that opens downward with x-intercepts at 1 and 3.

- Focus-directrix form: The focus-directrix form of a parabola is given by the equation (x – h)² = 4p(y – k), where (h, k) is the vertex of the parabola, p is the distance between the vertex and the focus (or the vertex and the directrix), and the axis of symmetry is parallel to the y-axis. If p is positive, the focus is above the vertex, and if p is negative, the focus is below the vertex.

For example, the graph of the equation (x – 2)² = -8(y – 1) is a parabola that opens downward with vertex (2, 1), focus (2, -1), and directrix y = 3.

**Describe Ellipse**

An ellipse is a conic section that is formed by the intersection of a plane and a cone, where the plane is at an angle to the axis of the cone. The shape of the ellipse is determined by the distance between its two foci (plural of focus) and the length of its major and minor axes. The major axis is the longest distance between any two points on the ellipse, and it passes through the two foci. The minor axis is the shortest distance between any two points on the ellipse, and it is perpendicular to the major axis and bisects it at the center of the ellipse.

The equation of an ellipse in standard form is:

((x – h)² / a²) + ((y – k)² / b²) = 1

where (h, k) is the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis. The semi-major axis is half the length of the major axis, and the semi-minor axis is half the length of the minor axis.

The foci of the ellipse are located on the major axis, and their distance from the center is given by the equation:

c = √(a² – b²)

The eccentricity of the ellipse is a measure of how “stretched out” it is, and it is given by the equation:

e = c / a

The value of e ranges from 0 to 1, where e = 0 represents a circle and e = 1 represents a parabola. An ellipse with 0 < e < 1 is called an elliptical, and an ellipse with e = 1 is called a degenerate ellipse or a parabolic. An ellipse with e > 1 is called a hyperbola.

For example, the equation of the ellipse with center (3, 4), semi-major axis length 5, and semi-minor axis length 3 is:

((x – 3)² / 25) + ((y – 4)² / 9) = 1

The foci of the ellipse are located on the major axis, and their distance from the center is:

c = √(25 – 9) = 4

The eccentricity of the ellipse is:

e = 4 / 5 = 0.8

Therefore, this is an elliptical ellipse.

**Describe the standard equation of an Ellipse**

An ellipse is a type of conic section that is formed when a cone is intersected by a plane at an angle that is not perpendicular to the base of the cone. It can also be defined as a curve that is traced out by a point that moves in a plane in such a way that the sum of its distances from two fixed points (called the foci) is constant.

The standard equation of an ellipse is given by:

(x-h)^{2}/a^{2} + (y-k)^{2}/b^{2} = 1

where (h,k) is the center of the ellipse, a is the distance from the center to the vertices along the x-axis (also known as the semi-major axis), and b is the distance from the center to the vertices along the y-axis (also known as the semi-minor axis).

If a=b, then the ellipse is a special case called a circle, and its equation can be written as:

(x-h)^{2} + (y-k)^{2} = r^{2}

where (h,k) is the center of the circle and r is its radius.

For example, the equation of an ellipse centred at the origin with semi-major axis of 4 and semi-minor axis of 2 is:

x^{2}/16 + y^{2}/4 = 1

The foci of this ellipse are located at (0,±sqrt(12)), and its vertices are located at (±4,0).

**Recall the terms of the Ellipse: Major Axis, Minor Axis, and Latus Rectum**

The terms major axis, minor axis, and latus rectum are commonly used to describe the properties of an ellipse.

The major axis is the longest diameter of an ellipse, and it passes through the center of the ellipse. It is also called the principal axis or the transverse axis. The length of the major axis is 2a, where a is the semi-major axis of the ellipse.

The minor axis is the shortest diameter of an ellipse, and it also passes through the center of the ellipse. It is also called the conjugate axis or the conjugate diameter. The length of the minor axis is 2b, where b is the semi-minor axis of the ellipse.

The latus rectum is a line segment that passes through one of the foci of the ellipse and is perpendicular to the major axis. It is also known as the focal chord, and its length is 2c, where c is the distance from the center of the ellipse to one of its foci.

The relationship between the major axis, minor axis, and latus rectum can be expressed as:

c^{2} = a^{2} – b^{2}

For example, consider the equation of an ellipse given by:

(x – 2)^{2}/9 + (y + 1)^{2}/16 = 1

The center of this ellipse is at the point (2,-1), and its semi-major and semi-minor axes are 3 and 4, respectively. Using the formula above, we can find the distance from the center to one of the foci:

c^{2} = 3^{2} – 4^{2}

c^{2} = -7

c = sqrt(-7)i

Since the distance is imaginary, this ellipse does not intersect the x-axis or the y-axis. However, we can still find the length of the latus rectum using the formula:

l = 2c^{2}/a

l = 2(-7)/3

l = -14/3

So the length of the latus rectum of this ellipse is -14/3 units.

**Describe the Hyperbola**

The hyperbola is a conic section formed when a plane intersects a double cone in such a way that it cuts through both of the cones. There are two distinct types of hyperbolas: one where the two branches of the hyperbola are facing inwards towards each other, called a “real hyperbola,” and the other where the branches face away from each other, called an “imaginary hyperbola” or “degenerate hyperbola.”

The standard equation of a hyperbola is given by:

(x-h)^{2}/a^{2} – (y-k)^{2}/b^{2} = 1 (for a real hyperbola with horizontal transverse axis)

(y-k)^{2}/b^{2} – (x-h)^{2}/a^{2} = 1 (for a real hyperbola with vertical transverse axis)

(x-h)^{2}/a^{2} – (y-k)^{2}/b^{2} = -1 (for an imaginary hyperbola with horizontal transverse axis)

(y-k^{)2}/b^{2} – (x-^{h)2}/a^{2} = -1 (for an imaginary hyperbola with vertical transverse axis)

where (h, k) is the center of the hyperbola, “a” is the distance from the center to the vertices, “b” is the distance from the center to the co-vertices.

The standard hyperbola has two asymptotes, which are straight lines that the hyperbola gets closer and closer to but never touches. The equations of the asymptotes for a real hyperbola are given by:

y = ±(b/a)(x-h) + k

The distance between the two branches of a hyperbola is given by the “transverse axis” or “major axis.” For a real hyperbola with a horizontal transverse axis, the length of the transverse axis is 2a, while for a vertical transverse axis, it is 2b. The “conjugate axis” or “minor axis” is the line perpendicular to the transverse axis passing through the center of the hyperbola. The length of the conjugate axis is 2b for a horizontal transverse axis, and 2a for a vertical transverse axis.

The distance between a vertex and the corresponding focus is called the “focal distance,” denoted by c. The value of “c” can be calculated as:

c = sqrt(a^{2} + b^{2})

Hyperbolas also have a “latus rectum,” which is a line segment that passes through a focus and is perpendicular to the major axis. Its length is given by:

2l = 2b^{2}/a (for a real hyperbola with horizontal transverse axis)

2l = 2a^{2}/b (for a real hyperbola with vertical transverse axis)

Examples of hyperbolas include the trajectory of a planet orbiting the sun, and the shape of the dish reflector of a radio telescope.

**Recall the Basic Terminologies used in the Hyperbola**

The hyperbola is another type of conic section. It is defined as the set of all points in a plane, the difference of whose distances from two fixed points, called foci, is constant.

The basic terminologies used in hyperbola are:

- Center: The point of intersection of the axes of the hyperbola is called its center.
- Foci: The two fixed points inside the hyperbola, which determine its shape, are called foci. The distance between the foci is denoted by 2c.
- Transverse axis: The line passing through the foci and the center of the hyperbola is called the transverse axis. The length of the transverse axis is denoted by 2a.
- Conjugate axis: The line perpendicular to the transverse axis passing through the center is called the conjugate axis. The length of the conjugate axis is denoted by 2b.
- Vertices: The points of intersection of the transverse axis and the hyperbola are called the vertices.
- Asymptotes: The two lines passing through the center of the hyperbola such that the difference in the distance of any point on the hyperbola from these lines is constant are called asymptotes.
- Latus Rectum: The latus rectum of the hyperbola is the line segment passing through the foci, perpendicular to the transverse axis, and with its midpoint on the hyperbola. Its length is denoted by 2l.
- Eccentricity: The eccentricity of the hyperbola is the ratio of the distance between a focus and the center to the distance between a point on the hyperbola and the center. It is denoted by e.

The standard equation of a hyperbola is given by:

(x^{2} / a^{2}) – (y^{2} / b^{2}) = 1 for horizontal hyperbola

(y^{2} / b^{2}) – (x^{2} / a^{2}) = 1 for vertical hyperbola

where a is the distance between the center and vertex along the transverse axis, b is the distance between the center and vertex along the conjugate axis, and c is the distance between the center and focus.

For example, the equation of the horizontal hyperbola with center (0,0), transverse axis length 6, and conjugate axis length 4 is:

(x^{2} / 9) – (y^{2} / 4) = 1

The distance between the center and each focus is calculated as:

c^{2} = a^{2} + b^{2}

c^{2} = 9 + 4

c = √13

The foci are located at (±c, 0) = (±√13, 0).

**Recall standard equation of Hyperbola**

A hyperbola is a type of conic section that has two separate branches. The standard equation of a hyperbola is:

(x – h)²/a² – (y – k)²/b² = 1, for a hyperbola centred at (h, k) with horizontal transverse axis

or

(y – k)²/b² – (x – h)²/a² = 1, for a hyperbola centred at (h, k) with vertical transverse axis

where a is the distance from the center to each vertex along the transverse axis, b is the distance from the center to each co-vertex along the conjugate axis, and (h, k) is the center of the hyperbola.

The transverse axis is the axis that passes through the two vertices of the hyperbola, and it determines the shape of the hyperbola. If the transverse axis is horizontal, then the hyperbola is said to have a horizontal transverse axis, and if it is vertical, then the hyperbola is said to have a vertical transverse axis.

The foci of the hyperbola are located at a distance c from the center, where c is calculated using the formula:

c² = a² + b²

The asymptotes of a hyperbola are two straight lines that intersect at the center of the hyperbola and approach the branches of the hyperbola as they extend to infinity. The equations of the asymptotes are:

y – k = ± b/a(x – h), for a hyperbola centred at (h, k) with horizontal transverse axis

x – h = ± a/b(y – k), for a hyperbola centred at (h, k) with vertical transverse axis

Here are some examples of the standard equation of a hyperbola:

- (x – 2)²/16 – (y – 3)²/9 = 1 represents a hyperbola with center at (2, 3), transverse axis of length 8, and conjugate axis of length 6, with vertices at (6, 3) and (-2, 3) and foci at (4, 3) and (0, 3).
- (y + 1)²/9 – (x – 4)²/16 = 1 represents a hyperbola with center at (4, -1), transverse axis of length 6, and conjugate axis of length 8, with vertices at (4, 5) and (4, -7) and foci at (4, -4) and (4, 2).