The normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{9}=1$ at the point $(8,3\sqrt{3})$ on it passes through the point:

$$\begin{gathered} Given hyperbola \frac{x^2}{a^2}-\frac{y^2}{9}=1 \\ \sin ce the point (8,3\sqrt{3}) lie on the hyperbola then \end{gathered}$$ 

$\frac{64}{a^2}-\frac{27}{9}=1\Rightarrow a^2=\frac{64}{4}=16$

equation of normal at $(8,3\sqrt{3})$ is

$2\sqrt{3}x+3y-25\sqrt{3}=0$

verify options

option (c) is correct