Question:
Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a point on the curve $\mathrm{y}=\mathrm{x}^{2}+7 \mathrm{x}+2$, nearest to the line, $\mathrm{y}=3 \mathrm{x}-3$. Then the equation of the normal to the curve at $P$ is :
Correct Option: , 4
Solution:
Let $\mathrm{L}$ be the common normal to parabola $y=x^{2}+7 x+2$ and line $y=3 x-3$
$\Rightarrow$ slope of tangent of $y=x^{2}+7 x+2$ at $P=3$
$\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\text {For } \mathrm{P}}=3$
$\Rightarrow 2 x+7=3 \Rightarrow x=-2 \Rightarrow y=-8$
So $\mathrm{P}(-2,-8)$
Normal at $\mathrm{P}: \mathrm{x}+3 \mathrm{y}+\mathrm{C}=0$
$\Rightarrow \mathrm{C}=26$ (P satisfies the line)
Normal: $x+3 y+26=0$