Let H: -x^2 a^2 +y^2 b^2 =1 be the hyperbola, whose eccentricity … - Vidyadip

MCQ

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 :

A
$170$
✓
$171$
C
$169$
D
$172$

✓

Answer

Correct option: B.

$171$

b
$ \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} \quad \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=2 $

$ a^2=2 b^2 $

$ \text { length of L.R. }=\frac{2 a^2}{b}=4 \sqrt{3} $

$ \mathrm{a}=\sqrt{6} $

$ P(\alpha, 6) \text { lie on } \frac{y^2}{3}-\frac{x^2}{6}=1 $

$ 12-\frac{\alpha^2}{6}=1 \Rightarrow \alpha^2=66 $

$ \text { Foci }=(0, \pm \mathrm{be})=(0,3) \&(0,-3) $

Let $d_1 \& d_2$ be focal distances of $\mathrm{P}(\alpha, 6)$

$ 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{147 \times 75}=105 $

$ \alpha^2+\beta=66+105=171$

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