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

Question

Aicharya · Accepted Answer

$\begin{aligned} & f(x)=x^2+9 \quad g(x)=\frac{x}{x-9} \ & a=f(g(10))=f\left(\frac{10}{10-9}ight) \ & =f(10)=109 \ & b=g(f(3))=g(9+9) \ & =g(18)=\frac{18}{9}=2 \ & E: \frac{x^2}{109}+\frac{y^2}{2}=1\end{aligned}$ $\begin{aligned} & \mathrm{e}^2=1-\frac{2}{109}=\frac{107}{109} \ & \ell=\frac{2(2)}{\sqrt{109}}=\frac{4}{\sqrt{109}} \ & 8 \mathrm{e}^2+\ell^2=\frac{8(107)}{109}+\frac{16}{109} \ & =8\end{aligned}$

Let $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)$. If…

Solution