Question

The slope of the tangent to the curves $x=t^2+3t-8$, $y=2t^2-2t-5$ at the point $\left(2,-1\right)$ is

• A. $\frac{22}{7}$
• B. $\frac{6}{7}$
• C. $-6$
• D. $-7$

Answer 1

The correct option is B

$\frac{6}{7}$

Explanation for the correct answer:

Compute the slope:

$x=t^2+3t-8$ $...\left(i\right)$

Differentiate with respect to $t$

$\frac{dx}{dt}=2t+3$

$y=2t^2-2t-5$ $...\left(ii\right)$

Differentiate with respect to $t$

$\frac{dy}{dt}=4t-2$

By chain rule,

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

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

At $\left(2,-1\right)$

$t^2+3t-8=2$ (From $\left(i\right)$)

$\Rightarrow$ $t^2+3t-10=0$

$\Rightarrow$ $\left(t+5\right)\left(t-2\right)=0$

$\Rightarrow$ $t=-5$ or $t=2$ $...\left(iii\right)$

$2t^2-2t-5=-1$ (From $\left(ii\right)$)

$\Rightarrow$ $2t^2-2t-4=0$

$\Rightarrow$ $\left(t-2\right)\left(t+1\right)=0$

$\Rightarrow$ $t=2$ or $t=-1$ $...\left(iv\right)$

From $\left(iii\right),\left(iv\right)$ we get $t=2$

$\Rightarrow$ $\frac{dy}{dx}=\frac{4\left(2\right)-2}{2\left(2\right)+3}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{6}{7}$

$\frac{dy}{dx}$ is the slope of the tangent to the given curve at a point.

Hence, slope of the tangent to the given curve at $\left(2,-1\right)$ is $\frac{6}{7}$.

Hence, option $\left(B\right)$ is the correct answer.