Let f (x) = x^2 + 9, g(x) x / x - 9 and a = fog(10), b = gof(3) . If e and l denote the eccentricity and the length

Question

Let \(f(x)=x^{2}+9,\ g(x)=\frac{x}{x-9}\) and \(\mathrm{a}=\) fog(10), \(\mathrm{b}=\) 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.

(1) 16

(2) 8

(3) 6

(4) 12

Answer 1

Correct option is :  (2) 8 

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

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

\(=\mathrm{f}(10)=109\)

\(\mathrm{b}=\mathrm{g}\ (\mathrm{f}(3))=\mathrm{g}\ (9+9)\)

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

\(\mathrm{E}: \frac{\mathrm{x}^{2}}{109}+\frac{\mathrm{y}^{2}}{2}=1\)

\(\mathrm{e}^{2}=1-\frac{2}{109}=\frac{107}{109}\)

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

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

\(=8\)