Hyperbola Question - None

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 ($\alpha, 6$), $\alpha \gt 0$ lies on $H$. If $\beta$ is the product of the focal distances of the point ($\alpha, 6$), then $\alpha^{2}+\beta$ is equal to:

Accepted Answer

Equation of hyperbola
$H: \frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1, e=\sqrt{3}$
$e=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{3} \Rightarrow \frac{a^{2}}{b^{2}}=2$
length of latus rectum $=\frac{2 a^{2}}{b}=4 \sqrt{3}$
$\Rightarrow a=\sqrt{6}$
${P}(\alpha, 6)$ lie on $\frac{y^{2}}{3}-\frac{x^{2}}{6}=1$
$\Rightarrow 12-\frac{\alpha^{2}}{6}=1 \Rightarrow \alpha^{2}=66$
Foci $=(0, \pm b e)=(0,3) \&(0,-3)$
Let $d_{1} \& d_{2}$ be the focal distances of $P(\alpha$, 6)
$d_{1}=\sqrt{\alpha^{2}+(6+b e)^{2}}$
$d_{2}=\sqrt{\alpha^{2}+(6-b e)^{2}}$
$d_{1}=\sqrt{66+81}, d_{2}=\sqrt{66+9}$
$\beta=d_{1} d_{2}=\sqrt{144 imes 75}=105$
$\alpha^{2}+\beta=66+105=171$

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