JEE Main 2024 — Hyperbola Question with Solution

Let $\mathrm{H}:\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\left({\mathrm{b}}^2={\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)\right)$ $\begin{aligned} & \therefore \mathrm{eq}^{\mathrm{n}} \text { of } \mathrm{C}_1 \\ & \text { Ar. }=36 \pi \\ & \pi \mathrm{a}^2=36 \pi \\ & \mathrm{a}=6 \end{aligned}$ $\therefore {\mathrm{eq}}^{\mathrm{n}}\text{ of }{\mathrm{C}}_1={\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$
Now radius of ${\mathrm{C}}_2$ can be $\mathrm{a}(\mathrm{e}-1)$ or $\mathrm{a}(\mathrm{e}+1)$ for $\mathrm{r}=\mathrm{a}(\mathrm{e}-1)$ for $\mathrm{r}=\mathrm{a}(\mathrm{e}+1)$
Ar. $=4\pi$ $\pi r^2=4\pi$ $\pi {\mathrm{a}}^2(\mathrm{e}-1{)}^2=4\pi$ ${\mathrm{a}}^2(\mathrm{e}+1{)}^2=4$ $36\pi (e-1{)}^2=4\pi$ $36(e+1{)}^2=4$ $\mathrm{e}-1=\frac{1}{3}$ $\mathrm{e}+1=\frac{1}{3}$ $\mathrm{e}=\frac{4}{3}$ $-\frac{2}{3}$ Not possible $\begin{aligned} & \therefore \mathrm{b}^2=36\left(\frac{16}{9}-1\right)=28 \\ & \therefore L R=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 28}{6}=\frac{28}{3} \end{aligned}$