JEE Main 2022 — Ellipse Question with Solution

Given,

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

Now using eccentricity formula, $e^2=1-\frac{b^2}{a^2}$

We get, $\frac{1}{16}=1-\frac{b^2}{a^2}$

$\Rightarrow \frac{b^2}{a^2}=1-\frac{1}{16}=\frac{15}{16}\Rightarrow b^2=\frac{15}{16}{a}^2$

Now again $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is passing through $\left(-4\sqrt{\frac{2}{5}},3\right)$ on satisfying the point we get,

$\Rightarrow \frac{16\times \frac{2}{5}}{a^2}+\frac{9}{b^2}=1$

$\Rightarrow \frac{32}{5a^2}+\frac{9}{b^2}=1$

Now putting the value $b^2=\frac{15}{16}{a}^2$ in above equation we get,

$\frac{32}{5a^2}+\frac{9}{\frac{15}{16}{a}^2}=1$

$\frac{80}{5a^2}=1$

$16=a^2$

So, $b^2=15$ and $a^2+b^2=15+16=31$