Let be the hyperbola, whose eccentricity is and the length of the latus rectum is . Suppose the point (, 6), > 0 lies on H. If is the product of the focal distances of the point (, 6), then is equal to [2024]

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Solution

Q.
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 ( $α$ , 6), $α$ > 0 lies on H. If $β$ is the product of the focal distances of the point ( $α$ , 6), then $α^{2} + β$ is equal to [2024]

1 169

2 171

3 172

4 170

Ans.
(2)

$H : \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ and $e = \sqrt{3}$

Now, $e = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{3}$

$\Rightarrow \frac{a^{2}}{b^{2}} = 2 \Rightarrow a^{2} = 2 b^{2}$

Length of latus rectum $= \frac{2 a^{2}}{b} = \frac{4 b^{2}}{b} = 4 b = 4 \sqrt{3}$ [Given]

$\Rightarrow b = \sqrt{3} \Rightarrow a = \sqrt{6}$ [ $∵$ a > 0]

Now, P( $α$ , 6) lies on $H : \frac{y^{2}}{3} - \frac{x^{2}}{6} = 1$

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

Now, foci of hyperbola = (0, $\pm$ be) = (0, $\pm$ 3)

Let $d_{1}$ and $d_{2}$ be focal distancee of ( $α$ , 6)

$\Rightarrow d_{1} = \sqrt{α^{2} + {(6 - 3)}^{2}}, d_{2} = \sqrt{α^{2} + {(6 + 3)}^{2}}$

$\Rightarrow d_{1} = \sqrt{66 + 9}, d_{2} = \sqrt{66 + 81}$

$d_{1} = \sqrt{75}, d_{2} = \sqrt{147}$

Now, $β = d_{1} d_{2} = \sqrt{11025} = 105$

Now, $α^{2} + β = 66 + 105 = 171$ .

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