JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
The point $P(-2\sqrt{6},\sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $\frac{\sqrt{5}}{2}$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively, then $QR$ is equal to :
Hyperbola
Solution
$e^2=\tfrac54\Rightarrow b^2=\dfrac{a^2}{4}$. $P$ on the curve: $\dfrac{24}{a^2}-\dfrac{12}{a^2}=1\Rightarrow a^2=12$, $b^2=3$. Tangent at $P$: $\dfrac{-2\sqrt6x}{12}-\dfrac{\sqrt3y}{3}=1$; at $x=0$: $Q(0,-\sqrt3)$. Normal: $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$, i.e. $\dfrac{12x}{-2\sqrt6}+\sqrt3y=15$; at $x=0$: $R(0,5\sqrt3)$. $QR=6\sqrt3$.