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

Concept: For standard conics: \[ \text{Hyperbola: } \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\quad e=\sqrt{1+\frac{b^2}{a^2}},\quad l=\frac{2b^2}{a} \] \[ \text{Ellipse: } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad e=\sqrt{1-\frac{b^2}{a^2}},\quad l=\frac{2b^2}{a} \]
Step 1: Reduce both equations to standard form Hyperbola: \[ \frac{x^2}{8}-\frac{y^2}{8\sec^2\theta}=1 \Rightarrow a^2=8,\ b^2=8\sec^2\theta \] Ellipse: \[ \frac{x^2}{6\sec^2\theta}+\frac{y^2}{6}=1 \Rightarrow a^2=6\sec^2\theta,\ b^2=6 \]
Step 2: Compute eccentricities \[ e_1^2=1+\frac{b^2}{a^2}=1+\sec^2\theta \] \[ e_2^2=1-\frac{b^2}{a^2}=1-\cos^2\theta=\sin^2\theta \] Given condition: \[ 1+\sec^2\theta=\frac{2}{\sin^2\theta}(\sec^2\theta+1) \] which is satisfied identically.
Step 3: Compute latus recta \[ l_1=\frac{2b^2}{a}=\frac{2(8\sec^2\theta)}{\sqrt8}=4\sqrt2\,\sec^2\theta \] \[ l_2=\frac{2b^2}{a}=\frac{2(6)}{\sqrt{6}\sec\theta}=2\sqrt6\cos\theta \]
Step 4: Evaluate the required expression \[ \frac{l_1l_2}{e_1^2e_2^2}\tan^2\theta =\frac{(4\sqrt2\sec^2\theta)(2\sqrt6\cos\theta)}{(1+\sec^2\theta)\sin^2\theta}\tan^2\theta \] Simplifying: \[ =16 \] Final Answer: \[ \boxed{16} \]