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

Given that, \(x=t^{2}+3 t-8 \rightarrow(1)\)
\(y=2 t^{2}-2 t-5 \rightarrow(2)\)
At point \((2,-1)\), Eq. (2) becomes
\(-1=2 t^{2}-2 t-5\)
\(\Rightarrow 2 t^{2}-2 t-4=0\)
\(\Rightarrow t^{2}-t-2=0\)
\(\Rightarrow(t-2)(t+1)=0\)
\(\Rightarrow t=2,-1 \rightarrow(3)\)
Similarly, at point \((2,-1)\) Eq. (1) becomes
\(2=t^{2}+3 t-8\)
\(\Rightarrow t^{2}+3 t-10=0\)
\(\Rightarrow(t+5)(t-2)=0\)
\(\Rightarrow t=2,-5 \rightarrow(4)\)
From Eqs. (3) and (4), we have common value of \(t=2\)
Now, \(\frac{d y}{d t}=4 t-2\) and \(\frac{d x}{d t}=2 t+3\)
So, slope of the tangent to the curve is
\(\frac{d y}{d x}=\frac{4 t-2}{2 t+3}\)
\(\Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{4(2)-2}{(2)+3}=\frac{6}{7}\)