Question

The slope of the 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. $\frac{7}{6}$
• D. $\frac{-6}{7}$

Answer 1

Answer: (B)

We have, $x=t^2+3t-8$ and $y=2t^2-2t-5$
Put $x=2$
$\therefore 2=t^2+3t-8$
$\Rightarrow t^2+3t-10=0$
$\Rightarrow (t+5)(t-2)=0$
$\therefore t=-5,2$
Now, put $y=-1$
$\therefore -1=2t^2-2t-5$
$\Rightarrow 2t^2-2t-4=0$
$\Rightarrow t^2-t-2=0$
$\Rightarrow (t-2)(t+1)=0$
$\Rightarrow t=2,-1$
$\therefore t=2[\because t\ne -1]$
Now, $\frac{dx}{dt}=2t+3$ and $\frac{dy}{dt}=4t-2$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$
$=\frac{4t-2}{2t+3}$
$\therefore {\left(\frac{dy}{dx}\right)}_{t=2}=\frac{4(2)-2}{2(2)+3}$
$=\frac{8-2}{4+3}=\frac{6}{7}$
Hence, the slope of tangent is $\frac{6}{7}$.