Mathematics - Ellipse Question with Solution | TestHub

Equation of ellipse whose centre at $\left(0,0\right)$ is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

It passes through $(4,-1)$

$\Rightarrow \frac{16}{a^2}+\frac{1}{b^2}=1$

$\Rightarrow 16b^2+a^2=a^2{b}^2 ...\left(i\right)$

Again it passes through $(-2,2)$

$\Rightarrow \frac{4}{a^2}+\frac{4}{b^2}=1$

$\Rightarrow 4b^2+4a^2=a^2{b}^2 ...\left(ii\right)$

From the equation $\left(i\right)$ and $\left(ii\right)$, we get, 

$16b^2+a^2=4a^2+4b^2$

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

  Now, we have, $e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4}$

$\Rightarrow e=\frac{\sqrt{3}}{2}$.