JEE Main 2025 — 7 April, Evening Shift — Question 44

Equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Directrix: $x=\frac{-9}{5}$ and corresponding foci $(-5,0)$

$\Rightarrow \quad-\frac{a}{e}=-\frac{9}{5}$ and $-a e=-5$

$\Rightarrow \frac{9 e^{2}}{5}=5 \Rightarrow e=\sqrt{\frac{25}{9}}=\frac{5}{3} \Rightarrow a=3$

$\therefore \quad b^{2}=a^{2}\left(e^{2}-1\right)=9\left(\frac{25}{9}-1\right)=16$

Hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$

$(\alpha, 2 \sqrt{5})$ lie on it

$\Rightarrow \frac{\alpha^{2}}{9}-\frac{20}{16}=1 \Rightarrow \alpha^{2}=\frac{36}{16} \times 9=\frac{81}{4}$

Product for distance of $\left(x_{1} y_{1}\right)$ from the two foci

$=\left(e x_{1}+a\right)\left|e x_{1}-a\right|$

$=e^{2} x_{1}^{2}-a^{2}$

For $(\alpha, 2 \sqrt{5}) \Rightarrow P=\frac{25}{9} \cdot \frac{81}{4}-9=\frac{189}{4}$

$4 P=189$