JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

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

\(\begin{array}{l}\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\end{array} \)
.

3. If the coordinates of the vertices is (0, ±a) then use the equation

\(\begin{array}{l}\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\end{array} \)
.

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

Related Links:

Ellipse and Hyperbola – JEE Important and Previous Year Questions

Frequently Asked Questions

Q1

Give the definition of an ellipse.

We can define an ellipse as a set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.

Q2

Define eccentricity of an ellipse.

We define eccentricity of an ellipse as the ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse.

Q3

Give the equation to find the eccentricity of an ellipse.

The eccentricity of an ellipse is given by the formula, e = √(1 – (b2/a2)).

Q4

How to find the equation of ellipse, when vertices and eccentricity is given?

Using the equation, e = c/a, find the value of c. If the vertices are (±a, 0), use (x2/a2) + (y2/b2) = 1. If the vertices are (0, ±a), use the equation, (x2/b2) + (y2/a2) = 1. Find b2 using c2 = a2-b2. Substitute the values of b2 and a2 in the standard equation.

Find the equation of ellipse with vertices and eccentricity

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