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
$x=t^{2}+3 t-8$
$\therefore \, \frac{d x}{d t}=2 t+3$
and $ y=2 t^{2}-2 t-5$
$\therefore \, \frac{d y}{d t}=4 t-2$
Slope of tangent $=\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{4 t-2}{2 t+3} \,\,\,\,\,\dots(i)$
Since, curve passes through the point $(2,-1)$.
$\therefore \, t^{2}+3 t-8=2 $
and $2 t^{2}-2 t-5=-1 $
$\Rightarrow \, t^{2}+3 t-10=0 $
and $2 t^{2}-2 t-4=0 $
$\Rightarrow \, t^{2}+5 t-2 t-10=0 $
and $ t^{2}-t-2=0 $
$\Rightarrow \,(t+5)(t-2)=0 $ and $\left(t^{2}-2 t+t-2\right)=0$
$\Rightarrow t=-5, 2$ 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
$\left[\frac{d y}{d x}\right]_{\text {at } t=2}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}$
$\therefore $ Slope of normal $=\frac{-1}{\frac{d y}{d x}}=\frac{-1}{(6 / 7)}=-\frac{7}{6}$