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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 :

(1) $x+3y-62=0$

(2) $x+3y+26=0$

(3) $x-3y-11=0$

(4) $x-3y+22=0$
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Ans. (2) $x+3y+26=0$

Sol.

Tangent at $p(h, k)$ will be parallel to given line

${dy\over dx}\Big|_{(h,k)}= 2h + 7 = 3\implies h=-2$

Point $P$ lies on curve $K=(-2)^2-7×2+2=-8$

Normal at $P(-2, -8)$, normal slop $=-{1\over3}$

$x+3y+26=0$

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Your diagrammatic solution makes clear concept on topic
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