The slope of the tangent to the curve

Given that $x=t^{2}+3 t-8, y=2 t^{2}-2 t-5$

Differentiating w.r.t. $t$,

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

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

For $(2,-1)$,

The given point is $(2,-1)$

$2=t^{2}+3 t-8,-1=2 t^{2}-2 t-5$

On solving we get,

$\mathrm{t}=2$ or $-5$ and $\mathrm{t}=2$ or $-1$

$\because \mathrm{t}=2$ is the common solution

$\mathrm{SO}, \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8-2}{4+3}=\frac{6}{7}$