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: Option: 1 $x+3y+26=0$ Option: 2 $x+3y-26=0$ Option: 3 $x-3y-11=0$   Option: 4 $x-3y+22=0$

Let L be the common normal to the parabola

$y=x^{2}+7 x+2 \text { and line } y=3 x-3$

\begin{aligned} &\Rightarrow \text { slope of tangent of } y=x^{2}+7 x+2 \text { 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\\ &\text { So } \mathrm{P}(-2,-8) \end{aligned}

Normal at P : x + 3y + C = 0

⇒C = 26 (P satifies the line)

Normal : x + 3y + 26 = 0

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