Explanation:
(B) : Given,
Curve \(x=t^{2}+3 t-8\)
\(y=2 t^{2}-2 t-5 \tag{i}\)
On differentiating w.r.t. t, we get-
\(\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t}+3\)
Curve \(y=2 t^{2}-2 t-5\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{t}-2\)
Then,
\(\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y}{d t} \times \frac{d t}{d x} \\ \frac{d y}{d x} & =\frac{4 t-2}{2 t+3}\end{aligned}\)
Given point is \((2,-1)\)
On putting the value of \(x\) in equation (i), we get-
\(2=\mathrm{t}^{2}+3 \mathrm{t}-8\)
\(\mathrm{t}^{2}+3 \mathrm{t}-10=0\)
\((\mathrm{t}-2)(\mathrm{t}+5)=0\)
\(\mathrm{t}=2, \mathrm{t}=-5\)
On putting the value of \(y=-1\) in equation (ii), we get -
\(-1=2 \mathrm{t}^{2}-2 \mathrm{t}-5 \Rightarrow 2 \mathrm{t}^{2}-2 \mathrm{t}-4=0\)
\(2\left(\mathrm{t}^{2}-\mathrm{t}-2\right)=0 \Rightarrow \mathrm{t}^{2}-\mathrm{t}-2=0\)
\((\mathrm{t}-2)(\mathrm{t}+1)=0 \Rightarrow \mathrm{t}=2, \mathrm{t}=-1\)
The common value of \(t\) is 2
Hence, The slope of the tangent to the given curve at point \((2,-1)\) is
Putting the common value \(t=2\) in equation (iii), we get
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4 \mathrm{t}-2}{2 \mathrm{t}+3}=\frac{4 \times(2)-2}{2(2)+3}=\frac{8-2}{7}\)
So, \(\quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{7}\)
