Let the ellipse E: x^2 144+ y^2 169=1 and the hyperbola H: x^2 16- y^2 λ^2=-1 have the same foci. If e and L respectively denote the eccentricity and…
JEE Main 2026 — Mathematics Coordinate Geometry
2026mcqmedium
Let the ellipse E:144x2+169y2=1 and the hyperbola H:16x2−λ2y2=−1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24(e+L) is :
Official previous-year question
Held on 28 Jan 2026 · Verified 6 Jul 2026.
Options
A
148
B
126
C
67
D
296
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Solution
Ellipse E: 144x2+169y2=1 has a2=169,b2=144, so cE=25=5 with foci at (0,±5).
Hyperbola H: x2y2−16x2=1 has the same foci, so cH2=x2+16=25, giving x2=9.
Thus H is 9y2−16x2=1 with a=3,b=4,c=5.
Eccentricity: e=35.
Latus rectum: L=a2b2=332.
Therefore 24(e+L)=24(35+332)=24⋅337=296
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