Question

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

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

Answer 1

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

$x=t^2+3t-8$ ........$\left(1\right)$

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

Differentiate $\left(1\right)$, we get

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

Differentiate $\left(2\right)$, we get

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

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

Given point is $(2,-1)$

Put the point in original x and y, we get

$\Rightarrow x=t^2+3t-8$

$2=t^2+3t-8$

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

$(t-2)(t+5)=0$

$t=2,t=-5$

$\Rightarrow y=2t^2-2t-5$

$(-1)=2t^2-2t-5$

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

$(t+1)(t-2)=0$

$t=-1,t=2$

Since, $t=2$ common in both parts, so we take

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

At $t=2$

$\frac{dy}{dx}=\frac{4\left(2\right)-2}{2\left(2\right)-3}=\frac{8-2}{4+3}=\frac{6}{7}$

$m=\frac{dy}{dx}=\frac{6}{7}$.