Let the domain of the function f(x)= _3 _5 _7(9x-x^2-13) be the interval (m,n).

Question

Let the domain of the function $f(x)={\log }_{3}{\log }_{5}{\log }_{7}(9x-{x}^{2}-13)$ be the interval $(m,n)$. Let the hyperbola $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ have eccentricity $\frac{n}{3}$ and the length of the latus rectum $\frac{8m}{3}$. Then ${b}^{2}-{a}^{2}$ is equal to :

Accepted Answer

$\begin{gathered} {\log }_{5}({\log }_{7}(9x-{x}^{2}-13))>0 \ \Rightarrow \, 9x-{x}^{2}-13>7 \ \Rightarrow \, x\in (4,5)=(m,n) \ \Rightarrow \, =\frac{5}{3} \ L(LR)=\frac{32}{3} \ \frac{25}{9}=1+\frac{{b}^{2}}{{a}^{2}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(1) \end{gathered}$

$\begin{gathered} \frac{2{b}^{2}}{a}=\frac{32}{3}\Rightarrow {b}^{2}=\frac{16a}{3} \ \ \frac{16}{9}=\frac{16a}{3{a}^{2}}\Rightarrow a=3 \ {b}^{2}=16 \ 	herefore {b}^{2}-{a}^{2}=16-9 \ =7 \end{gathered}$

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