Question

For $0 < \theta < \frac{pi}{2}$, if the eccentricity of the hyperbola $x^2 - y^2 \csc^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \csc^2 \theta + y^2 = 5$, then the value of $\theta$ is:

• A. $\frac{\pi}{6}$
• B. $\frac{5\pi}{12}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (C)

$\frac{\pi}{3}$

Answer 2

For the hyperbola, the eccentricity is given by:

$e_h = \sqrt{1 + \sin^2 \theta}$

For the ellipse, the eccentricity is given by:

$e_e = \sqrt{1 - \sin^2 \theta}$

We are given that the eccentricity of the hyperbola is $\sqrt{7}$ times the eccentricity of the ellipse:

$e_h = \sqrt{7} \cdot e_e$

Substituting the expressions for $e_h$ and $e_e$:

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

Squaring both sides:

$1 + \sin^2 \theta = 7(1 - \sin^2 \theta)$

Expanding:

$1 + \sin^2 \theta = 7 - 7 \sin^2 \theta$

Simplifying:

$1 + \sin^2 \theta + 7 \sin^2 \theta = 7$

$8 \sin^2 \theta = 6$

$\sin^2 \theta = \frac{3}{4}$

Thus:

$\sin \theta = \frac{\sqrt{3}}{2}$

Therefore:

$\theta = \frac{\pi}{3}$