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

$x+3y+26=0$

Explanation for the correct answer:

Step:1 Finding the coordinates of point $P$

Given: Curve $y=x^2+7x+2$, line $y=3x-3.$

Let $L$ be the common normal to parabola $y=x^2+7x+2$and line $y=3x–3$

Therefore slope of tangent of $y=x^2+7x+2$ at $P=3$

Differentiate w,r. to x

$$\begin{gathered} \Rightarrow 2x+7=3 \\ \Rightarrow 2x=-4 \\ \Rightarrow x=–2 \end{gathered}$$

Substituting value of $x$ in $y=x^2+7x+2$ ,we get ;

$$\begin{gathered} y=x^2+7x+2 \\ \Rightarrow y=4-14+2 \\ \Rightarrow y=-8 \end{gathered}$$

Step:2 Finding Slope of Normal

Since slope of tangent at P is $3$ and Normal and tangents areperpendicular to each other so, If m is the slope of normal then

$$\begin{gathered} m\times 3=-1 \\ m=\frac{-1}{3} \end{gathered}$$ since product of slope of two perpendicular lines is minus one.

Step:3 Finding equation of Normal

General equation of line is $y=mx+c$ since slope of normal is $m=\frac{-1}{3}$ and it passes through point P $(-2,-8)$ so,

$$\begin{gathered} -8=-2\left(\frac{-1}{3}\right)+c \\ c=\frac{-26}{3} \end{gathered}$$

So, Equation of normal becomes

$x+3y+26=0$

Hence the correct option is (D).