If the eccentricity of the hyperbola $x^2 - y^2\sec^2 a = 5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2\sec^2 a + y^2 = 25$, then a value of $a$ is:
Step-by-Step Solution
Key Concept: The relationship between eccentricities of hyperbola and ellipse is connected through the parameter $\alpha$ via trigonometric identities.
For the hyperbola $\frac{x^2}{5} - \frac{y^2}{5\cos^2\alpha} = 1$, the eccentricity is $e_1^2 = 1 + \cos^2\alpha$. For the ellipse $\frac{x^2}{25\cos^2\alpha} + \frac{y^2}{25} = 1$, the eccentricity is $e_2^2 = \sin^2\alpha$. Given $e_1 = \sqrt{3}e_2$, we have $e_1^2 = 3e_2^2$, which gives $1 + \cos^2\alpha = 3\sin^2\alpha$. Solving yields $\sin\alpha = \frac{1}{\sqrt{2}}$.
Correct Answer: 2