JEE Main 2021 (Online) 18th March Evening ShiftQuestion 1
Consider a hyperbola $H:x^{2}-2y^{2}=4$. Let the tangent at a point $P(4,\sqrt{6})$ meet the $x$-axis at $Q$ and latus rectum at $R(x_1,y_1)$, $x_1>0$. If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\triangle QFR$ is equal to :
Hyperbola
Solution
$H:\dfrac{x^2}{4}-\dfrac{y^2}{2}=1$, foci $(\pm\sqrt6,0)$, latus rectum $x=\sqrt6$. Tangent at $P(4,\sqrt6)$: $4x-2\sqrt6\,y=4$. With $y=0$: $Q(1,0)$. At $x=\sqrt6$: $y_1=2-\dfrac{\sqrt6}{3}$. Nearer focus $F(\sqrt6,0)$. Area $=\tfrac12\cdot QF\cdot y_1=\tfrac12(\sqrt6-1)\left(2-\dfrac{\sqrt6}{3}\right)=\dfrac{7}{\sqrt6}-2$.