Explanation

(a)
Given, $y = mx + \frac{25\sqrt{3}}{3}\ \ \ \ \ldots\ (i)$
and $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1\ \ \ \ \ \ \ \ \ \ \ \ldots\ (ii)$
Here, Eq. (i) is normal to Eq. (ii), then
$\frac{\left( a^{2} + b^{2} \right)^{2}}{c^{2}} = \frac{a^{2}}{m^{2}} - \frac{b^{2}}{1}$
$\Rightarrow \ \ \frac{(16 + 9)^{2} \times 9}{625 \times 3} = \frac{16}{m^{2}} - \frac{9}{1}$
$\Rightarrow \frac{16}{m^{2}} = 12\ \Rightarrow m = \pm \frac{2}{\sqrt{3}}$
$\left\lbrack \begin{array}{r}
\therefore condition\ for\ \ lx + my + n \\
= 0\ to\ be\ a\ hyperbola\ is\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{\left( a^{2} + b^{2} \right)^{2}}{n^{2}}
\end{array} \right\rbrack$