The slope of the tangent to the curve $x = {t^2} + 3t - 8,y = 2{t^2} - 2t - 5$ at the point $(2, - 1)$ is

(A) $\cfrac{{22}}{7}$

(B) $\cfrac{6}{7}$

(C)$\cfrac{7}{6}$

(D)$\cfrac{{ - 6}}{7}$

Option B is correct

We have, $x = {t^2} + 3t - 8\;andy = 2{t^2} - 2t - 5$ …(i)

At $(2,\; - 1),2 = {t^2} + 3t - 8 \Rightarrow {t^2} + 3t - 10 = 0$

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

and $- 1 = 2{t^2} - 2t - 5 \Rightarrow 2{t^2} - 2t - 4 = 0 \supset 2(t - 2)(t + 1) = 0$

$\Rightarrow t = 2, - 1$

Thus, at $x = 2$ and $y = - 1,t = 2.$

Differentiating (i) w.r.t. $t,\cfrac{{dx}}{{dt}} = 2t + 3$ and

$\cfrac{{dv}}{{dt}} = 4t - 2$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}}$

$= \cfrac{{4t - 2}}{{2t + 3}} \& \therefore \,\,\,\,\,{\left( {\cfrac{{dy}}{{dx}}} \right)_{r = 2}} = \cfrac{{4(2) - 2}}{{2(2) + 3}} = \cfrac{6}{7}$