JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Hyperbola

The latus rectum gives $\frac{2b^2}{a}=9$. The tangent $y-\sqrt{3}x+\sqrt{3}=0$ has slope $\sqrt{3}$ and intercept $-\sqrt{3}$, and for $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ the tangency condition $c^2=a^2{m}^2-b^2$ becomes $3=3a^2-b^2$. Solving these together gives $a=2$ and $b^2=9$, so the curve is $\frac{x^2}{4}-\frac{y^2}{9}=1$ with $e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$. The point of contact is $\left(-\frac{a^{2}m}{c},-\frac{b^2}{c}\right)=(4,3\sqrt{3})$, so $x_0=4$. For a point on a hyperbola the two focal distances are $e{x}_0-a$ and $e{x}_0+a$, whose product is $m=e^2{x}_0^2-a^2$. Hence $4e^2+m=e^2(4+x_0^2)-a^2=20e^2-a^2=20\cdot \frac{13}{4}-4=61$. The directrix printed as $x=\pm \frac{4}{\sqrt{3}}$ was inconsistent with $e>1$; the correct directrix is $x=\pm \frac{a}{e}=\pm \frac{4}{\sqrt{13}}$.