Let the hyperbola H: x^2 a^2-y^2=1 and the ellipse E:3x^2+4y^2=12 be such that the length of latus rectum of H is equal to the length of latus rectum…
JEE Main 2022 — Mathematics Coordinate Geometry
2022integermedium
Let the hyperbola H:a2x2−y2=1 and the ellipse E:3x2+4y2=12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of 12(eH2+eE2) is equal to _____.
Official previous-year question
Held on 24 Jun 2022 · Verified 6 Jul 2026.
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Solution
Given
E≡3x2+4y2=12⇒4x2+3y2=1
Now, eE=1−a2b2=1−43=21
Length of L.R. =a2b2=22×3=3
Now H≡a2x2−1y2=1
Length of L.R.=a2b2=a2×1
Given length of L.R. of E and H are equal
So a2=3⇒a=32
Now eH=a2b2+1=(32)21+1⇒eH=49+1=413
So 12(eH2+eE2)=12(413+41)=12(414)=14×3=42
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