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 $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ have eccentricity $\dfrac{n}{3}$ and length of latus rectum $\dfrac{32}{3}$. Then $b^2-a^2$ is equal to:

Accepted Answer

For $f$ to be defined, $\log_7(9x-x^2-13)>1 \Rightarrow 9x-x^2-13>7 \Rightarrow x^2-9x+20<0 \Rightarrow x\in(4,5)$. So $(m,n)=(4,5)$.

Eccentricity $e=\dfrac{n}{3}=\dfrac53$, so $e^2=\dfrac{25}{9}=1+\dfrac{b^2}{a^2}\ \ (1)$.

Latus rectum: $\dfrac{2b^2}{a}=\dfrac{32}{3}\Rightarrow b^2=\dfrac{16a}{3}$.

Substituting into (1): $\dfrac{16}{9}=\dfrac{16}{3a}\Rightarrow a=3$, so $b^2=16$.

$b^2-a^2=16-9=7.$

Solutions · 1