Let f(x)=x^2+9,g(x)=(x)/(x-9) and a=f g(10),b=g f(3).

Question

Let $f(x)={x}^{2}+9,g(x)=\frac{x}{x-9}$ and $a=f\circ g(10),b=g\circ f(3)$. If $e$ 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 $8{e}^{2}+{l}^{2}$ is equal to.

Accepted Answer

$\begin{gathered} g(10)=10 \ a=f(g(10))=f(10)=109 \ f(3)=18 \ b=g(f(3))=g(18)=2 \ \frac{{x}^{2}}{109}+\frac{{y}^{2}}{2}=1 \ e=\sqrt{1-\frac{2}{109}}=\sqrt{\frac{107}{109}} \ I=\frac{2{b}^{2}}{a}=\frac{2	imes 2}{\sqrt{109}} \end{gathered}$

$\begin{gathered} 8{e}^{2}+{l}^{2} =\frac{8	imes 107}{109}+\frac{16}{109} \ =8 \end{gathered}$

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