# FIND CENTER VERTICES AND CO VERTICES OF AN ELLIPSE

## About "Find center vertices and co vertices of an ellipse"

Find center vertices and co vertices of an ellipse :

What is an ellipse ?

The locus of a point in a plane whose distance from a fixed point bears a constant ratio, less than one to its distance from a fixed line is called ellipse.

 Equation of ellipse symmetric about x-axis (where a > b) Equation of ellipse symmetric about y-axis (where a > b) x²/a² + y²/b² = 1 (center is (0, 0))(x-h)²/a² + (y-k)²/b² = 1(center is (h, k)) x²/b² + y²/a² = 1(center is (0, 0))(x-h)²/b² + (y-k)²/a² = 1(center is (h, k))

center :

The point of intersection of the major axis and minor axis of the ellipse is called the centre of the ellipse.

Here C(0, 0) is the centre of the ellipse.

Note that the centre need not be the origin of the ellipse always.

vertices :

The points of intersection of the ellipse and its major axis are called its vertices. Here the vertices of the ellipse are A(a, 0) and A′(− a, 0).

Co-vertices are B(0,b) and B'(0, -b).

## Find center vertices and co vertices of an ellipse - Examples

Example 1 :

Find the center, vertices and co-vertices of the following ellipse

x²/25 + y²/9  =  1

Solution :

From the given equation we come to know the number which is at the denominator of x is greater, so the ellipse is symmetric about x-axis.

Center :

In the above equation no number is added or subtracted with x and y. So the center of the ellipse is C (0, 0)

Vertices :

a² = 25 and b² = 9

a = 5 and b = 3

Vertices are A(a, 0) and A'(-a, 0)

A (5, 0) and A' (-5, 0)

Co-vertices :

B (0, b) and B' (0, -b)

B (0, 3) and B' (0, -3)

## How to find the center of an ellipse from standard form ?

Example 2 :

Find the center, vertices and co-vertices of the following ellipse

[(x - 1)²/9] + [(y + 1)²/16]  =  1

Solution :

From the given equation we come to know the number which is at the denominator of y is greater, so the ellipse is symmetric about y-axis.

Center :

In the above equation is in the form of

[(x - h)²/b²] + [(y - k)²/a²]  =  1

Here (h,k) is the center. Instead of "h" we have 1 and instead of "k" we have -1..

Hence center C (h, k) ==> (1, -1)

To find the vertices of the ellipse, we have to make some changes in the given equation.

Let X = x -1 and Y = y + 1

Vertices :

a² = 16 ==>  a = 4. Vertices of ellipse with center (0, 0) are A(a, 0) and A'(-a, 0)

A (4, 0) and A' (-4, 0)

Vertices of given ellipse :

 x - 1 = 4   y + 1 = 0x = 4 + 1   y = -1x = 5 and y = -1 x - 1 = -4   y + 1 = 0x = -4 + 1   y = -1x = -3 and y = -1

Hence the vertices of required ellipse are A(5, -1) and A'(-3, -1)

Co-vertices :

B (0, b) and B' (0, -b)

B (0, 3) and B' (0, -3)

 x - 1 = 0   y + 1 = 3x =  1   y = 3 - 1x = 1 and y = 2 x - 1 = 0   y + 1 = -3x =  1   y = -3 - 1x = 1 and y = -4

Hence the vertices of required ellipse are B(1, 2) and B'(1, -4)

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