Question

Let $P(h,k)$ be a point on the curve $y=x^2+7x+2,$ nearest to the line, $y=3x-3$. Then the equation of the normal to the curve at $P$ is:

• A. $x+3y-62=0$
• B. $x-3y-11=0$
• C. $x-3y+22=0$
• D. $x+3y+26=0$

Answer 1

The correct option is D $x+3y+26=0$

$C:y=x^2+7x+2$
Let P : $(h,k)$ lies on Curve$k=h^2+7h+2\cdots \left(1\right)$

Now for the shortest distance
Slope of tangent line at point $P=$ slope of line $L$
${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{\text{at P}(h,k)}$
$\Rightarrow {\begin{array}{c}\frac{d}{dx}(x^2+7x+2)\end{array}|}_{\text{at P}(h,k)}=3$
$\Rightarrow {\begin{array}{c}(2x+7)\end{array}|}_{\text{at P}(h,k)}=32h+7=3$
$h=-2$ from equation $\left(1\right)$
$k=-8$
$P:(-2,-8)$ equation of normal to the curve is perpendicular to
L : $3x-y=3$
$N:x+3y=\lambda$ pass through $(-2,-8)$
$\Rightarrow \lambda =-26$
$\therefore N:x+3y+26=0$