Question

Find the equation of ellipse whose eccentricity is $\frac{2}{3}$ , latus rectum is $5$ and the centre is $(0,0)$.

Answer 1

Given: Eccentricity $e=\frac{2}{3}$

Length of the latus rectum$=5$

Centre is $(0,0)$

$e=\frac{\sqrt{a^2-b^2}}{a}=\frac{2}{3}$

$\Rightarrow 9a^2-9b^2=4a^2$

$\Rightarrow 5a^2=9b^2$

$\Rightarrow b^2=\frac{5a^2}{9}$ .......$\left(1\right)$

Given:$\frac{2b^2}{a}=5$

$\Rightarrow b^2=\frac{5a}{2}$ ........$\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$ we get

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

$\Rightarrow \frac{1}{2}=\frac{a}{9}$

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

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

$\therefore b^2=\frac{5a}{2}=\frac{5\times \frac{9}{2}}{2}=\frac{45}{4}$

$\therefore$ The required equation of the ellipse is $\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$

or $\frac{4x^2}{81}+\frac{4y^2}{45}=1$ is the required equation.