Find the equation of an ellipse whose eccentricity is

Question:

Find the equation of an ellipse whose eccentricity is $\frac{2}{3}$, the latus rectum is 5, and the center is at the origin.

Solution:

Let the equation of the required ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ …(i)

Given that

Eccentricity $=\frac{2}{3}$

we know that,

Eccentricity, $e=\frac{c}{a}$

$\Rightarrow \frac{2}{3}=\frac{c}{a}$

$\Rightarrow c=\frac{2}{3} a$

We know that,

$c^{2}=a^{2}-b^{2}$

$\Rightarrow\left(\frac{2 a}{3}\right)^{2}=a^{2}-b^{2}$

$\Rightarrow \frac{4 a^{2}}{9}=a^{2}-b^{2}$

$\Rightarrow b^{2}=a^{2}-\frac{4 a^{2}}{9}$

$\Rightarrow b^{2}=\frac{9 a^{2}-4 a^{2}}{9}$

$\Rightarrow \mathrm{b}^{2}=\frac{5 \mathrm{a}^{2}}{9}$ …(ii)

It is also given that, Latus Rectum = 5 …(iii)

We know that,

Latus Rectum $=\frac{2 b^{2}}{a}$

$\Rightarrow 5=\frac{2 \times\left(\frac{5 a^{2}}{9}\right)}{a}$

$\Rightarrow 5=\frac{10 a^{2}}{9 a}$

$\Rightarrow 5=\frac{10 a}{9}$

$\Rightarrow a=\frac{5 \times 9}{10}$

$\Rightarrow a=\frac{9}{2}$

$\Rightarrow a^{2}=\frac{81}{4}$

Substituting the value of a in eq. (ii), we get

$b^{2}=\frac{5\left(\frac{9}{2}\right)^{2}}{9}$

$\Rightarrow \mathrm{b}^{2}=\frac{5 \times 9}{4}$

$\Rightarrow \mathrm{b}^{2}=\frac{45}{4}$

Substituting the value of $a^{2}$ and $b^{2}$ in eq. (i), we get

$\frac{x^{2}}{\frac{81}{4}}+\frac{y^{2}}{\frac{45}{4}}=1$

$\Rightarrow \frac{4 x^{2}}{81}+\frac{4 y^{2}}{45}=1$