For 0 < < /2, if the eccentricity of the hyperbola is times eccentricity of the ellipse , then the value of is [2024]

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Solution

Q.
For 0 < $θ$ < $π$ /2, if the eccentricity of the hyperbola $x^{2} - y^{2} {cosec}^{2} θ = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^{2} {cosec}^{2} θ + y^{2} = 5$ , then the value of $θ$ is [2024]

1 $π / 6$

2 $π / 3$

3 $5 π / 12$

4 $π / 4$

Ans.
(2)

Given, $\frac{x^{2}}{5} - \frac{y^{2}}{5 \sin^{2} θ} = 1$

$a = \sqrt{5} > b = \sqrt{5} \sin θ$

$e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \sin^{2} θ}$ ... (i)

Also, $\frac{x^{2}}{5 \sin^{2} θ} + \frac{y^{2}}{5} = 1$ (Given)

$a = \sqrt{5} \sin θ < b = \sqrt{5}$

$e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \sin^{2} θ}$ ... (ii)

Given, eqn. (i) = $\sqrt{7}$ eqn. (ii)

$\sqrt{1 + \sin^{2} θ} = \sqrt{7} \sqrt{1 - \sin^{2} θ}$

Squaring on both sides.

$1 + \sin^{2} θ = 7 (1 - \sin^{2} θ) \Rightarrow 1 + \sin^{2} θ = 7 - 7 \sin^{2} θ$

$\Rightarrow 8 \sin^{2} θ = 6$

$\sin^{2} θ = \frac{6}{8} = \frac{3}{4} \Rightarrow \sin θ = \frac{\sqrt{3}}{2}$

So, $θ = \frac{π}{3}$ .

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