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∈(0,2π), let the eccentricity and the length of the latus rectum of the hyperbola x2−y2sec2θ=8 be e1 and l1, respectively, and let the eccentricity and the length of the latus rectum of the ellipse x2sec2θ+y2=6 be e2 and l2, respectively. If e12=e22(sec2θ+1), then (e1e2l1l2)tan2θ is equal to ____
Official previous-year question
Held on 28 Jan 2026 · Verified 6 Jul 2026.
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Solution
8x2−8cos2θy2=1, e1=1+88cos2θ
ℓ1=a2b2=222⋅(8cos2θ)
6x2+6cos2θy2=1; e2=1−66cos2θ=sinθ
ℓ2=a2b2=62⋅6cos2θ
e12=e22(1+sec2θ)
1+cos2θ=sin2θ(1+cos2θ1)
1+cos2θ=sin2θ+tan2θ
Solving we get θ=4π
ℓ1=22
e1=23
ℓ2=6
e2=21
(e1e2ℓ1ℓ2)tan2θ=8 (By putting values)
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