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…

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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

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{H}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1, \mathrm{e}=\sqrt{3} \\ & \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{3} \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=2 \\ & \qquad \mathrm{a}^2=2 \mathrm{~b}^2 \\ & \text { length of L.R. }=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=4 \sqrt{3} \\ & \qquad a=\sqrt{6}\end{aligned}$ $P(\alpha, 6)$ lie on $\frac{y^2}{3}-\frac{x^2}{6}=1$ $\begin{aligned} & 12-\frac{\alpha^2}{6}=1 \Rightarrow \alpha^2=66 \\ & \text { Foci }=(0, \pm \text { be })=(0,3) \&(0,-3) \end{aligned}$ Let $d_1 \& d_2$ be focal distances of $\mathrm{P}(\alpha, 6)$ $\begin{aligned} & \mathrm{d}_1=\sqrt{\alpha^2+(6+b e)^2}, \mathrm{~d}_2=\sqrt{\alpha^2+(6-b e)^2} \\ & \mathrm{~d}_1=\sqrt{66+81}, \mathrm{~d}_2=\sqrt{66+9} \\ & \beta=\mathrm{d}_1 \mathrm{~d}_2=\sqrt{147 \times 75}=105 \\ & \alpha^2+\beta=66+105=171\end{aligned}$
SECTION - B

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

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