Question

Find the equation of the ellipse referred to its centre whose latus rectum is $5$ and whose eccentricity is $\frac{2}{3}$.

Answer 1

General Equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Given : $e=\frac{2}{3}$ and Latus rectum $=5$

We know that, Latus rectum $=\frac{2b^2}{a}$
$\frac{2b^2}{a}=5$ ..... [Given]

$\Rightarrow b^2=\frac{5a}{2}$ .... $\left(i\right)$
Also, $b^2=a^2(1-e^2)$

$=a^2(1-\frac{4}{9})$ ....... $[\because e=\frac{2}{3}]$
$\Rightarrow \frac{5a}{2}=a^2\left(\frac{5}{9}\right)$ ..... From $\left(i\right)$
$\Rightarrow a=\frac{9}{2}$
From $\left(i\right)$, we get

$b^2=\frac{5}{2}\times \frac{9}{2}=\frac{45}{4}$
Therefore equation of ellipse becomes :
$\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$

$\Rightarrow 20x^2+36y^2=405$