JEE Main 2024 — 6 April, Shift 2 — Question 22

Given $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=9$ and $\frac{\mathrm{a}}{\mathrm{e}}= \pm \frac{4}{\sqrt{3}}$

equation of tangent $y-\sqrt{3} x+\sqrt{3}=0$

by equation of tangent

Let slope $=\mathrm{S}=\sqrt{3}$

Constant $=-\sqrt{3}$

By condition of tangency

$\Rightarrow 6=6 \mathrm{a}^{2}-9 \mathrm{a}$

$\Rightarrow \mathrm{a}=2, \mathrm{~b}^{2}=9$

Equation of Hyperbola is

$\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$ and for tangent

Point of contact is $(4,3 \sqrt{3})=\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$

Now $\mathrm{e}=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$

Again product of focal distances

$$\begin{aligned} & \mathrm{m}=\left(\mathrm{x}_{0} \mathrm{e}+\mathrm{a}\right)\left(\mathrm{x}_{0} \mathrm{e}-\mathrm{a}\right) & \mathrm{m}+4 \mathrm{e}^{2}=20 \mathrm{e}^{2}-\mathrm{a}^{2} & \quad=20 \times \frac{13}{4}-4=61 \end{aligned}$$

(There is a printing mistake in the equation of directrix $x= \pm \frac{4}{\sqrt{3}}$.

Corrected equation is $\mathrm{x}= \pm \frac{4}{\sqrt{13}}$ for directrix, as eccentricity must be greater than one, so question must be bonus)