Mathematics · 2020
JEE Main · 7 January 2020, Shift 2 · Q66
If 3 x+4 y=12 √ 2 is a tangent to the ellipse x^2/(a^2)+y^2/9=1 for some a ∈ R, then the distance between the foci of the ellipse is:
If $\displaystyle 3 x+4 y=12 \sqrt{2}$ is a tangent to the ellipse $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{9}=1$ for some $\displaystyle \mathrm{a} \in \mathbf{R}$, then the distance between the foci of the ellipse is :
Official answer
(1)
$\displaystyle 2 \sqrt{7}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.