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

$\textbf{Step 1 : Convert the given equation of hyperbola in parametric form}$ $Given : \,16x^2-9y^2=144$

$\implies \,\cfrac{16x^2}{144}-\cfrac{9y^2}{144}=1$

$\implies \,\cfrac{x^2}{9}-\cfrac{y^2}{16}=1\,\qquad\qquad...(i)$

$\implies a^2=9\,\implies a=3\quad ,b^2=16\qquad\qquad ...(ii)$

$\textbf{Step 2 : Write equation of directrix in standard form}$

$\textbf{Equation of directrix to the hyperbola }\,\boldsymbol{\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1 \,\textbf{is}}$

$\boldsymbol{x\pm\cfrac a e=0}$

$\text{Given equation of a directrix is}$

$5x+9=0$ $\text{dividing by 5 on both sides,}$ $\implies x+\cfrac 9 5=0$

$\because a=3,\qquad\qquad...[from\,eq^n \,(ii)]$

$\implies x+\cfrac {3} {\cfrac 5 3}=0$

$\therefore e=\cfrac 5 3 \,\text{ and directrix lies in left side of hyperbola.}$ $\text{Hence corresponding focus will be at negative }X-axis.$ $\therefore \textbf{focus is at } \boldsymbol{(-ae,0)}\equiv (-3)\times \cfrac 53\equiv (-5, 0)$

$\therefore \textbf{The focus corresponding to the directrix }\,\boldsymbol{5x+9=0\,\textbf{is at }\,(-5, 0) .}$

$\textbf{Hence, option C is correct .}$