10-2. Parabola, Ellipse, Hyperbola
hard

જો  $\mathrm{e}_{1}$ અને  $\mathrm{e}_{2}$ એ અનુક્રમે ઉપવલય $\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1$ અને અતિવલય $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$ ની  ઉકેન્દ્રીતા હોય  અને બિંદુ $\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ એ ઉપવલય $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},$ પર હોય તો  $\mathrm{k}$ મેળવો.

A

$15$

B

$14$

C

$17$

D

$16$

(JEE MAIN-2020)

Solution

$e_{1}=\sqrt{1-\frac{4}{18}}=\frac{\sqrt{7}}{3}$

$\mathrm{e}_{2}=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}$

$\because \quad\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ lies on $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k}$

$\Rightarrow \quad 15 \mathrm{e}_{1}^{2}+3 \mathrm{e}_{2}^{2}=\mathrm{k}$

$\Rightarrow \quad k=16$

Standard 11
Mathematics

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