If the hyperbola $x^2/a^2-y^2/b^2=1(a>b)$ has eccentricity and the length  of the latus rectum respectively equal to $5/4$  and 9 then $ab =$

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ length of latus rectum }=9\Rightarrow \frac{2b^2}{a}=9\Rightarrow 2b^2=9a\to \left(1\right)$

Eccentricity $=\frac{5}{4},b^2=a^2\left(e^2-1\right)\Rightarrow b^2=a^2\left(\frac{25}{16}-1\right)\Rightarrow b^2=\frac{9}{16}{a}^2\to \left(2\right)$

From(l) &(2); $2\left(\frac{9}{16}{a}^2\right)=9a\Rightarrow a=8.\text{ From }\left(1\right);2b^2=9\times 8\Rightarrow b^2=36\Rightarrow b=6$