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Hyperbola

Section: Mathematics

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Question : 25 of 98

Marks: +1, -0

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

H

\beta

is the product of the focal distances of the point

(\alpha, 6)

\alpha^{2}+\beta

[8 Apr 2024 Shift 1]

170
171
169
172

Solution: 👈: Video Solution

\; 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

\; a^{2}=2 b^{2}

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

\; a=\sqrt{6}

P(\alpha, 6)

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 b e)=(0,3) \& (0,-3)

Let

d_{1} \& d_{2}

be 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{147 \times 75}=105

\alpha^{2}+\beta=66+105=171

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