Explanation

(a)
Given, $\frac{x^{2}}{32/5} + \frac{y^{2}}{32/9} = 1$
Let the equation of tangent be $y = mx + c$
$y = mx \pm \sqrt{\frac{32}{5}m^{2} + \frac{32}{9}}\ \ \ \ \ \ldots(i)$
$\left\lbrack \therefore c^{2} = a^{2}m^{2} + b^{2}for\ a > b \right\rbrack$
Since, $(2,3)\ lies\ on\ Eq.(i)$
$\Rightarrow \ 3 = m.2 \pm \sqrt{\frac{32}{5}m^{2} + \frac{32}{9}}$
$45(3 - 2m)^{2} = 288m^{2} + 160$
$\Rightarrow 108m^{2} + 540m - 245 = 0$
$\therefore D = (540)^{2} + 4.180.245 > 0\ \Rightarrow \ D > 0$
$\Rightarrow$ Two values of$\ m\$will exist
$\Rightarrow$ Two tangents will exist
Alternate
Let $S \equiv 5x^{2} + 9y^{2} - 32$
Now, $S(2,3) \equiv 20 + 81 - 32 > 0$
$\therefore$ Point$\$(2,3) lies outside ellipse
Thus, two tangents can be drawn