Mathematics · 2022
JEE Main · 25 June 2022, Shift 2 · Q20
The line y=x+1 meets the ellipse x^2/4+y^2/2=1 at two points P and Q. If r is the radius of the circle with P Q as diameter then (3 r)^2 is equal to:
The line $\displaystyle y=x+1$ meets the ellipse $\displaystyle \frac{x^2}{4}+\frac{y^2}{2}=1$ at two points $\displaystyle P$ and $\displaystyle Q$. If $\displaystyle r$ is the radius of the circle with $\displaystyle P Q$ as diameter then $\displaystyle (3 r)^2$ is equal to :
Official answer
(1)
$\displaystyle 20$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.