Mathematics · 2021
JEE Main · 31 August 2021, Shift 2 · Q66
An angle of intersection of the curves, x^2/(a^2)+y^2/(b^2)=1 and x^2+y^2= ab, a > b, is:
An angle of intersection of the curves, $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ and $\displaystyle x^2+y^2=\mathrm{ab}, \mathrm{a}>\mathrm{b}$, is :
Official answer
(1)
$\displaystyle \tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.