For $0 < \theta < \frac{\pi}{2}$, if the eccentricity of the hyperbola $x^{2} - y^{2} \cosec^{2} \theta = 5$…

For $0 < \theta < \frac{\pi}{2}$, if the eccentricity of the hyperbola $x^{2} - y^{2} \cosec^{2} \theta = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^{2} \cosec^{2} \theta + y^{2} = 5$, then the value of $\theta$ is:
  1. π6
  2. 5π12
  3. π3
  4. π4

Solution

Given: The equation of hyperbola is, $x^{2} - y^{2} \csc^{2}\theta = 5$ $\Rightarrow \frac{x^{2}}{5} - \frac{y^{2}}{5\sin^{2}\theta} = 1$ And the equation of ellipse is $x^{2}\csc^{2}\theta + y^{2} = 5$ $\Rightarrow \frac{x^{2}}{5\sin^{2}\theta} + \frac{y^{2}}{5} = 1$ Here, $a < b$ as $\sin^{2}\theta \leq 1$ So, $e_{H} = \sqrt{1 + \sin^{2}\theta}$ and $e_{E} = \sqrt{1 - \sin^{2}\theta}$ Also given, $e_{H} = \sqrt{7}e_{E}$ $\Rightarrow \sqrt{1 + \sin^{2}\theta} = \sqrt{7}\sqrt{1 - \sin^{2}\theta}$ $\Rightarrow 1 + \sin^{2}\theta = 7 - 7\sin^{2}\theta$ $\Rightarrow 8\sin^{2}\theta = 6$ $\Rightarrow \sin\theta = \frac{\sqrt{3}}{2}$ $\Rightarrow \theta = \frac{\pi}{3}$

Asked in: JEE Main 2024 (01 Feb Shift 1)

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