Question

Find the equation of ellipse whose eccentricity is $\frac{1}{2}$ , latus rectum is $2$ and the centre is $(0,0)$.

Answer 1

Given: Eccentricity $e=\frac{1}{2}$

Length of the latus rectum$=2$

Centre is $(0,0)$

$e=\frac{\sqrt{a^2-b^2}}{a}=\frac{1}{2}$

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

$\Rightarrow 3a^2=4b^2$

$\Rightarrow b^2=\frac{3a^2}{4}$ .......$\left(1\right)$

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

$\Rightarrow b^2=a$ ........$\left(2\right)$

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

$a=\frac{3a^2}{4}$

$\Rightarrow 1=\frac{3a}{4}$

$\Rightarrow a=\frac{4}{3}$

$\Rightarrow a^2=\frac{16}{9}$

$\therefore b^2=a=\frac{4}{3}$

$\therefore$ The required equation of the ellipse is $\frac{x^2}{\frac{16}{9}}+\frac{y^2}{\frac{4}{3}}=1$

or $\frac{9x^2}{16}+\frac{3y^2}{4}=1$ is the required equation.