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10-2. Parabola, Ellipse, Hyperbola
hard
Let $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=\mathrm{fog}(10), \mathrm{b}=\operatorname{gof}(3)$. If $\mathrm{e}$ and $1$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$, then $8 e^2+1^2$ is equal to.
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
