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JEE Main 2021
26-08-2021 S1
Question
On the ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$ let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $S^{\prime}$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle SPS' then, the value of $\left(5-e^2\right) . A$ is :
Select the correct option:
A
6
B
12
C
14
D
24
✓ 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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