Let H:-x^2a^2+y^2b^2=1 be the hyperbola, whose eccentricity is √3 and the length of the la…

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

Accepted Answer

$\begin{gathered} H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=-1 \ e=\sqrt{1+\frac{{a}^{2}}{{b}^{2}}}=\sqrt{3} \ \Rightarrow 1+\frac{{a}^{2}}{{b}^{2}}=3 \ \Rightarrow \frac{{a}^{2}}{{b}^{2}}=2\, \ \frac{2{a}^{2}}{b}=4\sqrt{3} \end{gathered}$

Using equation (1)

$\begin{gathered} \frac{4{b}^{2}}{b}=4\sqrt{3} \ \Rightarrow b=\sqrt{3} \ a=\sqrt{6} \ H:\frac{{x}^{2}}{6}-\frac{{y}^{2}}{3}=-1 \ \frac{{\alpha }^{2}}{6}-12=-1 \ \frac{{\alpha }^{2}}{6}=11 \ \begin{gathered} {\alpha }^{2}=66 \ :(0,bc)(0,-bc) \ (0,3),(0,-3) \end{gathered} \ \beta =\sqrt{{\alpha }^{2}+9}	imes \sqrt{{\alpha }^{2}+81} \ \beta =105 \ {\alpha }^{2}+\beta =66+105 \ =171 \end{gathered}$

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