Mathematics · 2021
JEE Main · 16 March 2021, Shift 2 · Q65
If the points of intersections of the ellipse x^2/16+y^2/(b^2)=1 and the circle x^2+y^2=4 b, b >4 lie on the curve y^2=3 x^2, then b is equal to:
If the points of intersections of the ellipse $\displaystyle \frac{x^2}{16}+\frac{y^2}{\mathrm{~b}^2}=1$ and the circle $\displaystyle x^2+y^2=4 \mathrm{~b}, \mathrm{~b}>4$ lie on the curve $\displaystyle y^2=3 x^2$, then b is equal to:
Official answer
(4)
$\displaystyle 12$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.