Mathematics · 2022
JEE Main · 26 June 2022, Shift 2 · Q11
If m is the slope of a common tangent to the curves x^2/16+y^2/9=1 and x^2+y^2=12, then 12 m^2 is equal to:
If $\displaystyle m$ is the slope of a common tangent to the curves $\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1$ and $\displaystyle x^2+y^2=12$, then $\displaystyle 12 m^2$ is equal to :
Official answer
(2)
$\displaystyle 9$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.