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

$\frac{6}{7}$

The correct answer is Option (4) → $\frac{6}{7}$

$\frac{dx}{dt}=2t+3$, $\frac{dy}{dt}=4t-2$

$\frac{dy}{dx}=\frac{4t-2}{2t+3}$

at $x=2$

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

$t^2+3t-10=0$

$t=2,-5$

so $t=2$ satisfies $y=-1$

so $\frac{dy}{dx}=\frac{4×2-2}{2×2+3}=\frac{6}{7}$