JEE PYQ: Hyperbola Question 3
Question 3 - 2021 (18 Mar Shift 2)
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:
(1) $4\sqrt{6}$
(2) $\sqrt{6} - 1$
(3) $\frac{7}{\sqrt{6}} - 2$
(4) $4\sqrt{6} - 1$
Show Answer
Answer: (3)
Solution
Hyperbola: $\frac{x^2}{4} - \frac{y^2}{2} = 1$, so $a^2 = 4$, $b^2 = 2$, $e = \sqrt{1+\frac{1}{2}} = \sqrt{\frac{3}{2}}$. Focus $F(\sqrt{6}, 0)$. Tangent at $P$: $\frac{4x}{4} - \frac{\sqrt{6},y}{2} = 1$, i.e. $x - \frac{\sqrt{6}}{2}y = 1$. Meets $x$-axis at $Q(1, 0)$. Latus rectum: $x = \sqrt{6}$, so $R\left(\sqrt{6}, \frac{2}{\sqrt{6}}(\sqrt{6}-1)\right)$. Area of $\triangle QFR = \frac{1}{2}(\sqrt{6}-1)\cdot\frac{2}{\sqrt{6}}(\sqrt{6}-1) = \frac{7}{\sqrt{6}} - 2$.
BETA
