10-2. Parabola, Ellipse, Hyperbola
hard

ધારો કે $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ અને $\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)$. જો $\mathrm{e}$ અને $l$ એ ઉપવલય $\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$ ની અનુક્રમે ઉત્કેન્દ્રતા અને નાભિલંબની લંબાઈ દર્શાવે, તો $8 \mathrm{e}^2+l^2=$.................

A

$16$

B

$8$

C

$6$

D

$12$

(JEE MAIN-2024)

Solution

$ f(x)=x^2+9 \quad g(x)=\frac{x}{x-9} $

$ a=f(g(10))=f\left(\frac{10}{10-9}\right) $

$ =f(10)=109 $

$ b=g(f(3))=g(9+9) $

$ =g(18)=\frac{18}{9}=2 $

$ E: \frac{x^2}{109}+\frac{y^2}{2}=1 $

$ e^2=1-\frac{2}{109}=\frac{107}{109} $

$ \ell=\frac{2(2)}{\sqrt{109}}=\frac{4}{\sqrt{109}} $

$ 8 e^2+\ell^2=\frac{8(107)}{109}+\frac{16}{109} $

$ =8$

Standard 11
Mathematics

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