Question

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

• A. $\frac{6}{7}$
• B. $-\frac{6}{7}$
• C. $\frac{7}{6}$
• D. $-\frac{7}{6}$
• E. $-\frac{1}{2}$

Answer 1

Answer: (D)

Given curves are
$x=t^2+3t-8$
$\therefore \frac{dx}{dt}=2t+3$
and $y=2t^2-2t-5$
$\therefore \frac{dy}{dt}=4t-2$
Slope of tangent $=\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{4t-2}{2t+3}\dots (i)$
Since, curve passes through the point $(2,-1)$.
$\therefore t^2+3t-8=2$
and $2t^2-2t-5=-1$
$\Rightarrow t^2+3t-10=0$
and $2t^2-2t-4=0$
$\Rightarrow t^2+5t-2t-10=0$
and $t^2-t-2=0$
$\Rightarrow (t+5)(t-2)=0$ and $\left(t^2-2t+t-2\right)=0$
$\Rightarrow t=-5,2$ and $(t-2)(t+1)=0$
$\Rightarrow t=-5,2$ and $t=-1,2$
So, common value of $t$ is 2 .
On putting $t=2$ in Eq. (i), we get
${\left[\frac{dy}{dx}\right]}_{\text{at }t=2}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}$
$\therefore$ Slope of normal $=\frac{-1}{\frac{dy}{dx}}=\frac{-1}{(6/7)}=-\frac{7}{6}$