Question

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

• A. $\frac{22}{7}$
• B. $\frac{6}{7}$
• C. $-6$
• D. None of these

Answer 1

The correct option is B

$\frac{6}{7}$

Explanation for the correct option:

Find the slope of tangent to the given curve.

Given: curves are $x=t^2+3t–8$ … $\left(i\right)$ and

$y=2t^2-2t-5$ … $\left(ii\right)$

Since the curve passes through the point $\left(x,y\right)=(2,–1)$.

From equation $\left(i\right)$

$$\begin{gathered} 2=t^2+3t–8 \\ \Rightarrow t^2+3t–8-2=0 \\ \Rightarrow t^2+3t-10=0 \\ \Rightarrow t^2+5t-2t-10=0 \\ \Rightarrow t(t+5)-2(t+5)=0 \\ \Rightarrow (t-2)(t+5)=0 \\ \Rightarrow t=2,t=-5 \end{gathered}$$

From equation $\left(ii\right)$

$$\begin{gathered} -1=2t^2-2t-5 \\ \Rightarrow 2t^2-2t-4=0 \\ \Rightarrow t^2-t-2=0 \\ \Rightarrow t^2-2t+t-2=0 \\ \Rightarrow t\left(t-2\right)+1\left(t-2\right)=0 \\ \Rightarrow \left(t-2\right)\left(t+1\right)=0 \\ \Rightarrow t=2,t=-1 \end{gathered}$$

From both the solutions, we get $t=2$.
Differentiating the equation $\left(i\right)$ w.r.t. $t$, we get

$\frac{dx}{dt}=2t+3$ .…. $\left(iii\right)$

And differentiating the equation $\left(ii\right)$ w.r.t. $t$, we get

$\frac{dy}{dt}=4t-2$ ….. $\left(iv\right)$
Now, $\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}$

$=\frac{4t-2}{2t+3}$.... From $\left(iii\right)$ and $\left(iv\right)$

Therefore, the slope of tangent to the given curve ${\left(\frac{dy}{dx}\right)}_{\left(2,-1\right)}={\left(\frac{4t-2}{2t+3}\right)}_{t=2}$

$$\begin{gathered} =\frac{8-2}{4+3} \\ =\frac{6}{7} \end{gathered}$$

Hence, option $B$ is the correct answer.