# How to Find Equation of Ellipse with Vertices and Eccentricity

Ellipse can be defined as a set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. These fixed points are called foci of the ellipse. The major axis is the line segment which passes through the foci of the ellipse. The endpoints of this axis are called the vertices of the ellipse. The line segment which passes through the centre and perpendicular to the major axis is called the minor axis. In this article, we will learn how to find equation of ellipse with vertices and eccentricity.

The eccentricity of an ellipse is denoted by e. It is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse, i.e., e = c/a where a is the length of semi-major axis and c is the distance from centre to the foci.

## Steps to Find the Equation of the Ellipse With Vertices and Eccentricity.

1. Find c from equation e = c/a

2. If the coordinates of the vertices is (±a, 0) then use the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$.

3. If the coordinates of the vertices is (0, ±a) then use the equation $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1$.

4. Using the equation c2 = (a2 – b2), find b2.

5. Substitute the values of a2 and b2 in the standard form.

## Solved Examples

Example 1.

Find the equation of the ellipse, whose vertices are ( ± 9,0) and eccentricity ⅔.

Solution:

Given vertices are ( ± 9,0).

a = 9

eccentricity , e = c/a

⅔ = c/9

So c = 6

We know c2 = a2 – b2

So b2 = a2 – c2

= 92 – 62

= 81 – 36

= 45

We use the equation (x2/a2) + (y2/b2) = 1

Substitute a2 and b2

(x2/81) + (y2/45) = 1 is the required equation.

Example 2.

Find the equation of the ellipse, whose vertices are ( 0,± 20) and eccentricity 1/10.

Solution:

Given vertices are ( 0, ± 20).

a = 20

eccentricity , e = c/a

1/10 = c/20

So c = 2

We know c2 = a2 – b2

So b2 = a2 – c2

= 202 – 22

= 400 – 4

= 396

Here major axis is parallel to y axis.

We use the equation (x2/b2) + (y2/a2) = 1

Substitute a2 and b2

(x2/396) + (y2/400) = 1 is the required equation.

Example 3:

Find the equation of the ellipse with vertices ( ±10,0) and eccentricity ⅘.

Solution:

Given e = 4/5

a = 10

e = c/a

⅘ = c/10

c = 8

b2 = a2-c2

= 100-64

= 36

b = 6

Equation of ellipse is (x2/a2)+(y2/b2) = 1

(x2/100)+(y2/36) = 1