For some θ ∈ (0, π 2), if the eccentricity of the hyperbola, x^2-y^2 sec^2θ =10 is √5 times the eccentricity of the ellipse, x^2 sec^2θ +y^2=5, then…
JEE Main 2020 — Mathematics Coordinate Geometry
2020mcqmedium
For some θ∈(0,2π), if the eccentricity of the hyperbola, x2−y2sec2θ=10 is 5 times the eccentricity of the ellipse, x2sec2θ+y2=5, then the length of the latus rectum of the ellipse, is
Official previous-year question
Held on 2 Sept 2020 · Verified 6 Jul 2026.
Options
A
26
B
30
C
325
D
345
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Solution
For hyperbola 10x2−10cos2θy2=1
The eccentricity of the hyperbola eH=1+a2b2=1+cos2θ
For ellipse 5cos2θx2+5y2=1
The eccentricity of an ellipse eE=1−a2b2=1−cos2θ=sinθ
Given, eH=5eE
⇒1+cos2θ=5sin2θ⇒cos2θ=32
Now, the length of the latus rectum of an ellipse =a2b2=510cos2θ=3520=345
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