Hyperbola

Concept:The domain of a logarithmic function requires its argument to be positive. For a hyperbola, eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}}$ and latus rectum length is $\frac{2b^2}{a}$.Explanation:The function $f(x)=\log_3 \log_5 \log_7 (9x-x^2-13)$ is defined when $\log_5 \log_7 (9x-x^2-13) > 0$.Thus $\log_7 (9x-x^2-13) > 5^0 = 1$, so $9x-x^2-13 > 7^1 = 7$.This gives $9x - x^2 - 20 > 0$, or $x^2 - 9x + 20 < 0$.Factor: $(x-4)(x-5) < 0$, so domain is $(4,5)$. Hence $m=4$, $n=5$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$, eccentricity $e = \frac{n}{3} = \frac{5}{3}$ and latus rectum $LR = \frac{8m}{3} = \frac{32}{3}$.$e^2 = 1 + \frac{b^2}{a^2}$ gives $\left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{a^2}$, so $\frac{b^2}{a^2} = \frac{16}{9}$. Thus $b^2 = \frac{16}{9}a^2$. (1)Latus rectum: $\frac{2b^2}{a} = \frac{32}{3}$, so $b^2 = \frac{16}{3}a$. (2)Equate (1) and (2): $\frac{16}{9}a^2 = \frac{16}{3}a$. Cancel $16$: $\frac{a^2}{9} = \frac{a}{3}$.Multiply by $9$: $a^2 = 3a$, so $a(a-3)=0$. Since $a>0$, $a=3$.Then $b^2 = \frac{16}{3} \cdot 3 = 16$.Therefore $b^2 - a^2 = 16 - 9 = 7$.Answer:$7$, which corresponds to option A.