Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4 \sqrt{3}$. Suppose the point $(\alpha, 6), \alpha0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2+\beta$ is equal to [JEE Main 2024, 8 Apr (Shift 1)]

Question

Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4 \sqrt{3}$. Suppose the point $(\alpha, 6), \alpha>0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2+\beta$ is equal to [JEE Main 2024, 8 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: b) 171. $H:\frac{{y}^{2}}{{b}^{2}}-\frac{{x}^{2}}{{a}^{2}}=1,e=\sqrt{3}$ $e=\sqrt{1+\frac{{a}^{2}}{{b}^{2}}}=\sqrt{3}\Rightarrow \frac{{a}^{2}}{{b}^{2}}=2$ ${a}^{2}=2{b}^{2}$ length of L.R. $=\frac{2 a^2}{b}=4 \sqrt{3}$ $P(\alpha, 6)$ lie on $\frac{y^2}{3}-\frac{x^2}{6}=1$ $12-\frac{{\alpha }^{2}}{6}=1\Rightarrow {\alpha }^{2}=66$ Foci $=(0, \pm b e)=(0,3), (0,-3)$ Let ${d}_{1},{d}_{2}$ be focal distances of $P(\alpha ,6)$ ${d}_{1}=\sqrt{{\alpha }^{2}+(6+be{)}^{2}},{d}_{2}=\sqrt{{\alpha }^{2}+(6-be{)}^{2}}$ ${d}_{1}=\sqrt{66+81},{d}_{2}=\sqrt{66+9}$ $\beta ={d}_{1}{d}_{2}=\sqrt{147\times 75}=105$ ${\alpha }^{2}+\beta =66+105=171$

Let \(H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1\) be the hyperbola, whose eccentricity is \(\sqrt{3}\) and the length of the…

Practice more JEE Maths PYQs