JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Hyperbola

Given $e=\frac{\sqrt{5}}{2}$, so $e^2=\frac{5}{4}=1+\frac{b^2}{a^2}$, giving $\frac{b^2}{a^2}=\frac{1}{4}$, thus $b^2=\frac{a^2}{4}$. Point $P(-2\sqrt{6},\sqrt{3})$ lies on the hyperbola: $\frac{24}{a^2}-\frac{3}{b^2}=1$. Substituting $b^2=\frac{a^2}{4}$: $\frac{24}{a^2}-\frac{12}{a^2}=1\Rightarrow a^2=12$, $b^2=3$. The tangent at $(x_0,y_0)$ is $\frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1$. At $P(-2\sqrt{6},\sqrt{3})$: $\frac{-2\sqrt{6}x}{12}-\frac{\sqrt{3}y}{3}=1$. The conjugate axis is the $y$-axis ($x=0$), so $Q$ has $y$-coordinate $-\sqrt{3}$. The normal at $P$ is $\frac{a^{2}x}{x_0}+\frac{b^{2}y}{y_0}=a^2+b^2$. At $x=0$: $y=\frac{(a^2+b^2)y_0}{b^2}=\frac{15\sqrt{3}}{3}=5\sqrt{3}$. Thus $QR=|5\sqrt{3}-(-\sqrt{3})|=6\sqrt{3}$.