Explanation

(c)
The given equation can be rewritten as
$\frac{x^{2}}{20} + \frac{y^{2}}{\frac{45}{4}} = 1$
On comparing the given equation with the standard equation, we get
$a^{2} = 20,\ b^{2} = \frac{45}{4}$
$\therefore$ The equation of normal at the point $(2,\ 3)$ is
$\frac{x - 2}{\frac{2}{20}} = \frac{y - 3}{\left( \frac{12}{45} \right)}$
$\Rightarrow 40(x - 2) = 15(y - 3)$
$\Rightarrow 8x - 3y = 7 \Rightarrow 3y - 8x + 7 = 0$