JEE Main 2024 — Hyperbola Question with Solution

Given,

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

And its focii $\equiv \left(\pm 5, 0\right)$ and its latusrectum $\frac{2b^2}{a}=\sqrt{50}$

So, $ae=5$ and $b^2=\frac{5\sqrt{2}a}{2}$

Now, using the formula eccentricity we get,

$b^2=a^2\left(1-e^2\right)=\frac{5\sqrt{2}a}{2}$

$\Rightarrow a\left(1-e^2\right)=\frac{5\sqrt{2}}{2}$ $\left\{\text{as} a\ne 0\right\}$

$\Rightarrow \frac{5}{e}\left(1-e^2\right)=\frac{5\sqrt{2}}{2}$

$\Rightarrow \sqrt{2}-\sqrt{2}{e}^2=e$

$\Rightarrow \sqrt{2}{e}^2+e-\sqrt{2}=0$

$\Rightarrow \sqrt{2}{e}^2+2e-e-\sqrt{2}=0$

$\Rightarrow \sqrt{2}e\left(e+\sqrt{2}\right)-1\left(1+\sqrt{2}\right)=0$

$\Rightarrow \left(e+\sqrt{2}\right)\left(\sqrt{2}e-1\right)=0$

$\therefore e\ne -\sqrt{2}; e=\frac{1}{\sqrt{2}}$

$\Rightarrow a=5\sqrt{2}$ and $b=5$

Now, finding eccentricity of hyperbola,

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

We know that formula of eccentricity of hyperbola is given by, $B^2=A^2\left({e_H}^2-1\right)$ where $A^2=b^2 \& B^2=a^2{b}^2$

$\Rightarrow a^2{b}^2=b^2\left(e_H^2-1\right)$

$\Rightarrow a^2=\left(e_H^2-1\right)$

$\Rightarrow 50=\left(e_H^2-1\right)$

$\Rightarrow e_H^2=51$