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JEE MAIN 2026
28-01-2026 S2
Question
Let the ellipse $\mathrm{E}: \frac{x^2}{144}+\frac{y^2}{169}=1$ and the hyperbola $\mathrm{H}: \frac{x^2}{16}-\frac{y^2}{\lambda^2}=-1$ 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 $24(\mathrm{e}+\mathrm{L})$ is :
Select the correct option:
A
148
B
128
C
67
D
296
✓ Correct! Well done.
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Solution
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Question Tags
JEE Main
Mathematics
Easy
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