If $5x+9=0$ is the directrix of the hyperbola $16x^2-9y^2=144,$ then its corresponding focus is

The correct option is A $(-5,0)$
Given, equation of hyperbola $16x^2-9y^2=144$
$\Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1$

Here $a=3,b=4$
$e^2=1+\frac{16}{9}$
$\Rightarrow e=\frac{5}{3}$
Foci are $(\pm ae,0)\equiv (\pm 5,0)$
Now, given directrix is $x=-\frac{9}{5}$ which is of the form $x=-\frac{a}{e}$
Hence corresponding focus will be $(-ae,0)\equiv (-5,0)$