Question

If the eccentricity of the hyperbola \[ x^2 - y^2 \cos^2 \alpha = 25 \] is \( \sqrt{5} \), then the eccentricity of the ellipse \[ x^2 \cos^2 \alpha + y^2 = 5 \] is equal to:

• A. \( \sqrt{2} \)
• B. \( \sin^{-1} \frac{3}{4} \)
• C. \( \sin^{-1} \frac{\sqrt{5}}{4} \)
• D. None of these

Hint:

The eccentricity of a conic section is related to its parameters, and using the formulas for hyperbolas and ellipses allows you to compute the eccentricity.

Answer 1

The Correct Option is A

Using the relationship between the eccentricities of the hyperbola and ellipse, and knowing the given eccentricity of the hyperbola, we can calculate the eccentricity of the ellipse.
Final Answer: \[ \boxed{\sqrt{2}} \]