Question

Find the equation of the ellipse whose eccentricity is $\frac{2}{3}$, length of latus rectum is $5$, centre is $(0,0)$ and the major axis lies on $x-$axis

Answer 1

As the centre of the ellipse is $(0,0)$

i.e., origin and the major axis lies on $x-$axis, so its equation can be taken as

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .....$\left(1\right)$

According to given,$\frac{2b^2}{a}=5\Rightarrow b^2=\frac{5a}{2}$

Also, eccentricity$=\frac{c}{a}=\frac{2}{3}\Rightarrow c=\frac{2a}{3}$

We know that $c^2=a^2-b^2$

$\Rightarrow {\left(\frac{2a}{3}\right)}^2=a^2-\frac{5a}{2}$

$\Rightarrow \frac{4a^2}{9}-a^2=-\frac{5a}{2}$

$\Rightarrow \frac{4a^2-9a^2}{9}=\frac{-5a}{2}$

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

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

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

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

From $\left(1\right)$ equation of the ellipse is

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

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