Question

For some \(\theta \in \left(0, \frac{\pi}{2}\right)\), if the eccentricity of the hyperbola, \(x^2 - y^2\sec^2\theta = 10\) is \(\sqrt{5}\) times the eccentricity of the ellipse, \(x^2\sec^2\theta + y^2 = 5\), then the length of the latus rectum of the ellipse, is :

• A. \(\sqrt{30}\)
• B. \(2\sqrt{6}\)
• C. \(\frac{4\sqrt{5}}{3}\)
• D. \(\frac{2\sqrt{5}}{3}\)

Hint:

Convert both conics to standard form, write \(e_h^2\) and \(e_e^2\) in terms of \(\cos^2\theta\), use \(e_h^2 = 5e_e^2\) to get \(\cos^2\theta\), then use latus rectum \(2b^2/a\).

Answer 1

The Correct Option is C

Hyperbola: \(\frac{x^2}{10}-\frac{y^2}{10\cos^2\theta}=1\), so \(e_h^2 = 1+\frac{10\cos^2\theta}{10} = 1+\cos^2\theta\).
Ellipse: \(\frac{x^2}{5\cos^2\theta}+\frac{y^2}{5}=1\). Since \(5\cos^2\theta<5\), the major axis is along y with \(a^2=5,\ b^2=5\cos^2\theta\): \(e_e^2 = 1-\cos^2\theta\).
Given \(e_h = \sqrt5\,e_e\): \(1+\cos^2\theta = 5(1-\cos^2\theta)\Rightarrow 6\cos^2\theta = 4\Rightarrow\cos^2\theta=\frac23\).
Then \(b^2 = \frac{10}{3}\), \(a=\sqrt5\).
Latus rectum \(=\frac{2b^2}{a} = \frac{20}{3\sqrt5} = \frac{4\sqrt5}{3}\).
Answer: (3) \(\frac{4\sqrt{5}}{3}\)