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10-2. Parabola, Ellipse, Hyperbola
hard
If $3 x+4 y=12 \sqrt{2}$ is a tangent to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{9}=1$ for some a $\in \mathrm{R},$ then the distance between the foci of the ellipse is
A
$4$
B
$2\sqrt 7$
C
$2\sqrt 5$
D
$2\sqrt 2$
(JEE MAIN-2020)
Solution
$3 \mathrm{x}+4 \mathrm{y}=12 \sqrt{12}$ is tangent to $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{9}=1$
$c^{2}=m^{2} a^{2}+b^{2}$
$\Rightarrow a^{2}=16$
$\mathrm{e}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
Distance between focii $=2 \mathrm{ae}=2 \sqrt{7}$
Standard 11
Mathematics