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JEE MAIN 2024
31-01-2024 S1
Question
If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to
Select the correct option:
A
$7 \sqrt{\frac{2}{5}}-\frac{8}{3}$
B
$14 \sqrt{\frac{2}{5}}-\frac{4}{3}$
C
$14 \sqrt{\frac{2}{5}}-\frac{16}{3}$
D
$7 \sqrt{\frac{2}{5}}+\frac{8}{3}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Mathematics
Medium
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