The slope of the tangent to the curve

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

Differentiating both the sides,

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$\Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

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

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

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

$=\frac{6}{7}$