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

Answer: (B)

$\frac{6}{7}$

Given curves are $x=t^2+3t-8$ ... $\left(i\right)$ and $y=2t^2-2t-5$ ... $\left(ii\right)$

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

$t^2+3t-10=0⟹t=2$ or $t=-5$

From $\left(ii\right)$

$2t^2-2t-4=0⟹t^2-t-2=0⟹t=2$ or $t=-1$

From both the solutions, we get $t=2$

Differentiating both the equations w.r.t. $t$, we get

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

$\frac{dy}{dt}=4t-2$ ....... $\left(iv\right)$

Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

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

$\therefore \frac{dy}{dx}=\frac{4t-2}{2t+3}$ is the slope of tangent to the given curve

$\therefore {\left|\frac{dy}{dx}\right|}_{(2,-1)}={\left|\frac{4t-2}{2t+3}\right|}_{t=2}=\frac{8-2}{4+3}=\frac{6}{7}$ is the slope of tangent to the given curve at $(2,-1)$