JEE Main 2024 — Ellipse Question with Solution

Given: $A(\alpha ,0)$ and $B(0,\beta )$ lies on the line $5x+7y=50$.

$\Rightarrow 5\left(\alpha \right)+7\left(0\right)=50, 5\left(0\right)+7\left(\beta \right)=50$

$\Rightarrow \alpha =10, \beta =\frac{50}{7}$

Also, $P$ divide the line segment $AB$ internally in the ratio $7: 3$.

$\Rightarrow P\equiv \left(\frac{7\times 0+3\times 10}{7+3},\frac{7\times {\displaystyle \frac{50}{7}}+3\times 0}{7+3}\right)$

$\Rightarrow P\equiv \left(3,5\right)$

$\Rightarrow ae=3$ as perpendicular from $S$ passes through $P$

Now, $3x-25=0$ is a directrix of the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow x=\frac{25}{3}$

We know that, directrix of an ellipse is given by $x=\pm \frac{a}{e}$

$\Rightarrow \frac{a}{e}=\frac{25}{3}$

Also, $ae=3$

$\Rightarrow a=5, b=4$

Length of LR $=\frac{2b^2}{a}=\frac{32}{5}$