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Mathematics · 2026

JEE Main · 28 January 2026, Shift 1 · Q25

For some θ ∈(0, (π)/2), let the eccentricity and the length of the latus rectum of the hyperbola x^2-y^2 sec^2 θ=8 be e_1 and l_1, respectively, and…

For some $\displaystyle \theta \in\left(0, \frac{\pi}{2}\right)$, let the eccentricity and the length of the latus rectum of the hyperbola $\displaystyle x^2-y^2 \sec ^2 \theta=8$ be $\displaystyle e_1$ and $\displaystyle l_1$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $\displaystyle x^2 \sec ^2 \theta+y^2=6$ be $\displaystyle e_2$ and $\displaystyle l_2$, respectively. If $\displaystyle e_1^2=e_2^2\left(\sec ^2 \theta+1\right)$, then $\displaystyle \left(\frac{l_1 l_2}{e_1 e_2}\right) \tan ^2 \theta$ is equal to $\displaystyle \_\_\_\_$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.