10-2. Parabola, Ellipse, Hyperbola
hard

જો $3 x+4 y=12 \sqrt{2}$ એ કોઈક $a \in \mathrm{R},$ માટે ઉપવલય $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{9}=1$ નો સ્પર્શક હોય તો બંને નાભી વચ્ચેનું અંતર મેળવો.

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

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