Report Issue

JEE-Main 2024
01.02.24_(S1)
Question
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :
Select the correct option:
A
3
B
7/2
C
3/2
D
5/2
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Question Tags
JEE Main
Mathematics
Easy
Start Preparing for JEE with Competishun