Question

For some \( \theta\in\left(0,\frac{\pi}{2}\right) \), let the eccentricity and the length of the latus rectum of the hyperbola \[ x^2-y^2\sec^2\theta=8 \] be \( e_1 \) and \( l_1 \), respectively, and let the eccentricity and the length of the latus rectum of the ellipse \[ x^2\sec^2\theta+y^2=6 \] be \( e_2 \) and \( l_2 \), respectively. If \[ e_1^2=\frac{2}{e_2^2}\left(\sec^2\theta+1\right), \] then \[ \left(\frac{l_1l_2}{e_1^2e_2^2}\right)\tan^2\theta \] is equal to:

Hint:

Always convert conic equations to standard form before extracting parameters like eccentricity and latus rectum.

Answer 1

Correct Answer: 16

Step 1: Understanding the Concept:
Identify the standard equations of the given conics to find their eccentricities and lengths of latus rectum in terms of $\theta$.
Step 2: Key Formula or Approach:
Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $e^2 = 1 + \frac{b^2}{a^2}$, $LR = \frac{2b^2}{a}$.
Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $e^2 = 1 - \frac{\text{minor}^2}{\text{major}^2}$, $LR = \frac{2 \times \text{minor}^2}{\text{major}}$.
Step 3: Detailed Explanation:
Hyperbola: $\frac{x^2}{8} - \frac{y^2}{8\cos^2\theta} = 1 \implies a_1^2 = 8, b_1^2 = 8\cos^2\theta$.
$e_1^2 = 1 + \cos^2\theta$ and $l_1 = \frac{2(8\cos^2\theta)}{\sqrt{8}} = 4\sqrt{2}\cos^2\theta$.
Ellipse: $\frac{x^2}{6\cos^2\theta} + \frac{y^2}{6} = 1$. Since $\theta \in (0, \pi/2)$, $6\cos^2\theta<6$.
$e_2^2 = 1 - \cos^2\theta = \sin^2\theta$ and $l_2 = \frac{2(6\cos^2\theta)}{\sqrt{6}} = 2\sqrt{6}\cos^2\theta$.
Given: $e_1^2 = e_2^2(\sec^2\theta + 1) \implies 1 + \cos^2\theta = \sin^2\theta\left(\frac{1}{\cos^2\theta} + 1\right)$.
$1 + \cos^2\theta = \sin^2\theta \frac{1 + \cos^2\theta}{\cos^2\theta} \implies \tan^2\theta = 1 \implies \theta = \frac{\pi}{4}$.
At $\theta = \frac{\pi}{4}$: $\cos^2\theta = \frac{1}{2}, \sin^2\theta = \frac{1}{2}, e_1^2 = \frac{3}{2}, e_2^2 = \frac{1}{2}$.
$l_1 = 4\sqrt{2}(\frac{1}{2}) = 2\sqrt{2}$, $l_2 = 2\sqrt{6}(\frac{1}{2}) = \sqrt{6}$.
Target value: $\left(\frac{l_1 l_2}{e_1 e_2}\right) \tan^2\theta = \frac{(2\sqrt{2})(\sqrt{6})}{\sqrt{3/2} \cdot \sqrt{1/2}} \cdot 1 = \frac{2\sqrt{12}}{\sqrt{3}/2} = \frac{4\sqrt{3} \cdot 2}{\sqrt{3}} = 8$.
Step 4: Final Answer:
The result is 8.