Explanation

(a)
Given equation of curve is
$y^{2} = px^{3} + q$
$\therefore\ \ \ \ 2y\frac{dy}{dx} = 3px^{2}$
$\therefore\ \ \ \ \ \left( \frac{dy}{dx} \right)_{(2,\ 3)} = \frac{3p\ (2)^{2}}{2.3} = 2p$
The equation of tangent at (2, 3) is
$(y - 3) = 2p(x - 2)$
$\Rightarrow \ \ \ y = 2px - (4p - 3)$
This is similar to $y = 4x - 5$
$\therefore\ \ \ \ 2p = 4$ and $4p - 3 = 5$
$\Rightarrow \ p = 2$ and $p = 2$
Since, the point (2, 3) lies on the curve
$\therefore\ \ \ \ \ \ 9 = 8p + q\ \ \ \ \ \ \Rightarrow \ \ \ q = - 7$ [$\because p = 2$]