Question

Given conic \(x^2 - y^2 \sec^2 \theta = 8\) whose eccentricity is '\(e_1\)' & length of latus rectum '\(l_1\)' and for conic \(x^2 + y^2 \sec^2 \theta = 6\), eccentricity is '\(e_2\)' & length of latus rectum '\(l_2\)'. If \(e_1^2 = e_2^2 (1 + \sec^2 \theta)\) then value of \(\frac{e_1 l_1}{e_2 l_2} \tan \theta\)

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Simplify ratios early. In this problem, the \(\cos^2\theta\) factor in both latus rectums cancels out immediately, saving complex arithmetic with \(\theta\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves calculating the parameters (eccentricity and latus rectum) for a hyperbola and an ellipse, and finding a ratio under a specific geometric condition.
Step 2: Key Formula or Approach:
For \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) (Hyperbola): \( e = \sqrt{1 + \frac{b^2}{a^2}}, l = \frac{2b^2}{a} \).
For \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (Ellipse): \( e = \sqrt{1 - \frac{b^2}{a^2}}, l = \frac{2b^2}{a} \).
Step 3: Detailed Explanation:
Hyperbola 1: \( \frac{x^2}{8} - \frac{y^2}{8\cos^2\theta} = 1 \implies a_1^2=8, b_1^2=8\cos^2\theta \).
\( e_1^2 = 1 + \cos^2\theta \text{ and } l_1 = \frac{2(8\cos^2\theta)}{\sqrt{8}} = 4\sqrt{2}\cos^2\theta \).
Ellipse 2: \( \frac{x^2}{6} + \frac{y^2}{6\cos^2\theta} = 1 \implies a_2^2=6, b_2^2=6\cos^2\theta \).
\( e_2^2 = 1 - \cos^2\theta = \sin^2\theta \text{ and } l_2 = \frac{2(6\cos^2\theta)}{\sqrt{6}} = 2\sqrt{6}\cos^2\theta \).
Given Condition: \( e_1^2 = e_2^2(1 + \sec^2\theta) \)
\[ 1 + \cos^2\theta = \sin^2\theta(1 + \frac{1}{\cos^2\theta}) = \sin^2\theta + \tan^2\theta \]
\[ 1 - \sin^2\theta + \cos^2\theta = \tan^2\theta \implies 2\cos^2\theta = \tan^2\theta \implies \cos^2\theta = \frac{1}{2} \]
Calculating the ratio:
\[ \text{Ratio} = \frac{\sqrt{1+\cos^2\theta} \cdot 4\sqrt{2}\cos^2\theta}{\sin\theta \cdot 2\sqrt{6}\cos^2\theta} \cdot \tan\theta = \frac{2\sqrt{2} \cdot \sqrt{3/2}}{\sqrt{6} \cdot (1/\sqrt{2})} \cdot 1 = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \]
Step 4: Final Answer:
The value of the expression is 2.