Hyperbola
Hyperbola-Ellipse Pair — Eccentricity Condition
nta_pyq_2026_jan
Grade 11
Question:
For some $\theta\in\left(0,\dfrac{\pi}{2}\right)$, 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 $e_1^2=e_2^2(\sec^2\theta+1)$, then $\left(\dfrac{l_1 l_2}{e_1 e_2}\right)\tan^2\theta$ is equal to _____.
Step-by-Step Solution
Key Concept: Hyperbola: $\tfrac{x^2}{8}-\tfrac{y^2}{8\cos^2\theta}=1$. $a^2=8$, $b^2=8\cos^2\theta$. $e_1^2=1+\cos^2\theta$. $l_1=\tfrac{2b^2}{a}=\tfrac{16\cos^2\theta}{2\sqrt{2}}$. Ellipse: $\tfrac{x^2}{6\cos^2\theta}+\tfrac{y^2}{6}=1$ (major axis $y$ since $\cos^2\theta<1$). $e_2^2=1-\cos^2\theta/1$... use $e_2=\sin\theta$, $l_2=\tfrac{2\cdot6\cos^2\theta}{\sqrt{6}}$.
$\theta=\pi/4$. $\left(\tfrac{l_1l_2}{e_1e_2}\right)\tan^2\theta=8$.
Correct Answer: 8