JEE Main 2024 — Mathematics Coordinate Geometry

Let $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad\left(\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\right)$H:a2x2​− b2y2​=1( b2=a2(e2−1)) $\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 $\mathrm{C}_2$C2​ can be $\mathrm{a}(\mathrm{e}-1)$a(e−1) or $\mathrm{a}(\mathrm{e}+1)$a(e+1) for $\mathrm{r}=\mathrm{a}(\mathrm{e}-1)$r=a(e−1) for $\mathrm{r}=\mathrm{a}(\mathrm{e}+1)$r=a(e+1)

Ar. $=4 \pi$=4π $\pi r^2=4 \pi$πr2=4π $\pi \mathrm{a}^2(\mathrm{e}-1)^2=4 \pi$πa2(e−1)2=4π $\mathrm{a}^2(\mathrm{e}+1)^2=4$a2(e+1)2=4 $36 \pi(e-1)^2=4 \pi$36π(e−1)2=4π $36(e+1)^2=4$36(e+1)2=4 $\mathrm{e}-1=\frac{1}{3}$e−1=31​ $\mathrm{e}+1=\frac{1}{3}$e+1=31​ $\mathrm{e}=\frac{4}{3}$e=34​ $-\frac{2}{3}$−32​ Not possible $\begin{aligned}
& \therefore \mathrm{b}^2=36\left(\frac{16}{9}-1\right)=28 \
& \therefore L R=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 28}{6}=\frac{28}{3}
\end{aligned}$