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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…

JEE Main 2026 — Mathematics Coordinate Geometry

2026integermedium

For some θ∈(0,π2)\theta \in\left(0, \frac{\pi}{2}\right)θ∈(0,2π​), let the eccentricity and the length of the latus rectum of the hyperbola x2−y2sec⁡2θ=8x^{2}-y^{2} \sec ^{2} \theta=8x2−y2sec2θ=8 be e1e_{1}e1​ and l1l_{1}l1​, respectively, and let the eccentricity and the length of the latus rectum of the ellipse x2sec⁡2θ+y2=6x^{2} \sec ^{2} \theta+y^{2}=6x2sec2θ+y2=6 be e2e_{2}e2​ and l2l_{2}l2​, respectively. If e12=e22(sec⁡2θ+1)e_{1}^{2}=e_{2}^{2}\left(\sec ^{2} \theta+1\right)e12​=e22​(sec2θ+1), then (l1l2e1e2)tan⁡2θ\left(\frac{l_{1} l_{2}}{e_{1} e_{2}}\right) \tan ^{2} \theta(e1​e2​l1​l2​​)tan2θ is equal to ____\_\_\_\_____

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Held on 28 Jan 2026 · Verified 6 Jul 2026.

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