The slope of the tangent to the curve

The given curve is $x=t^{2}+3 t-8$ and $y=2 t^{2}-2 t-5$.

$\therefore \frac{d x}{d t}=2 t+3$ and $\frac{d y}{d t}=4 t-2$

$\therefore \frac{d y}{d x}=\frac{d y}{d t} \cdot \frac{d t}{d x}=\frac{4 t-2}{2 t+3}$

The given point is (2, −1).

At x = 2, we have:

$t^{2}+3 t-8=2$

$\Rightarrow t^{2}+3 t-10=0$

$\Rightarrow(t-2)(t+5)=0$

$\Rightarrow t=2$ or $t=-5$

At $y=-1$, we have:

$2 t^{2}-2 t-5=-1$

$\Rightarrow 2 t^{2}-2 t-4=0$

$\Rightarrow 2\left(t^{2}-t-2\right)=0$

$\Rightarrow(t-2)(t+1)=0$

$\Rightarrow t=2$ or $t=-1$

The common value of t is 2.

Hence, the slope of the tangent to the given curve at point (2, −1) is

$\left.\frac{d y}{d x}\right]_{t=2}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}$

The correct answer is B.