Question

State true or false:

If $x=t^2+3t-8,y=2t^2-2t-5$, then $\frac{dy}{dx}$ at $(2,-1)$ is $\frac{6}{7}$.

• A. True
• B. False

Answer 1

The correct option is A True

$x=t^2+3t-8\Rightarrow \frac{dx}{dt}=2t+3$
$y={2t}^2-2t-5\Rightarrow \frac{dy}{dt}=4t-2$
$\therefore \frac{dy}{dx}=\frac{4t-2}{2t+3}$
At $(2,-1)$
$2=t^2+3t-8\Rightarrow t^2+3t-10=0\Rightarrow t^2+5t-2t-10=0\Rightarrow (t+5)(t-2)=0\Rightarrow t=-5,t=2$
and
$-1={2t}^2-2t-5\Rightarrow {2t}^2-2t-4=0\Rightarrow t^2-t-2=0\Rightarrow t^2-2t+t-2=0\Rightarrow t=2,t=-1$
$\therefore t=2$
Hence $\frac{dy}{dx}=\frac{4.2-2}{2.2+3}=\frac{6}{7}$