Explanation

(b)
Given curve are
$x = t^{2} - 3t - 8,\ \ \ y = 2t^{2} - 2t - 5$ …(i)
When $x = 2$, then
$2 = t^{2} + 3t - 8$
$\Rightarrow \ \ \ t^{2} + 3t - 10 = 0\ \ \ \Rightarrow \ \ \ \ t = 2,\ - 5$ ...(ii)
When $y = - 1$, then
$- 1 = 2t^{2} - 2t - 5$
$\Rightarrow \ \ \ \ \ 2t^{2} - 2t - 4 = 0\ \ \ \ \Rightarrow \ \ \ \ t = - 1,\ 2$ ...(iii)
From Eqs. (ii) and (iii), $t = 2$
On differentiating Eq. (i) w.r.t. $t$ respectively, we get
$\frac{dx}{dt} = 2t + 3$ and $\frac{dy}{dt} = 4t - 2$
$\therefore\ \ \ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$
$\Rightarrow \ \ \ \ \left( \frac{dy}{dx} \right)_{(t = 2)} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$