Question

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

• A. $-6$
• B. $0$
• C. $\frac{6}{7}$
• D. $\frac{22}{7}$

Answer 1

The correct option is C $\frac{6}{7}$

$x=t^2+3t-8\Rightarrow \frac{dx}{dt}=2t+3$
$y=2t^2-2t-5\Rightarrow \frac{dy}{dt}=4t-2$
$\therefore \frac{dy}{dx}=\frac{4t-2}{2t+3}$
$\{\because (2,-1\left)\text{is a point on the curve}\right\}$
$\Rightarrow 2=t^2+3t-8$
$\Rightarrow t=-5,2$ and
$-1=2t^2-2t-5$ $\Rightarrow t=2,-1$
Thus $t=2$
$\Rightarrow$ Slope of tangent at the given point
$=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}$