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

The Correct Option is C

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:

\[ \theta = \frac{\pi}{3} \]