"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 $\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$, then $8 \mathrm{e}^2+l^2$ is equal to."

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Accepted Answer

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

$$ \begin{aligned} 8 e^2+l^2 & =\frac{8 \times 107}{109}+\frac{16}{109} \\ & =8 \end{aligned} $$

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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 $\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$, then $8 \mathrm{e}^2+l^2$ is equal to.

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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 $\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$, then $8 \mathrm{e}^2+l^2$ is equal to.

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