Let P(h, k) be a point on the curve y=x^{2}+7 | vedclass

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(D) The point $P$ on the curve $y=x^{2}+7x+2$ nearest to the line $y=3x-3$ is the point where the tangent to the curve is parallel to the given line.$

Vedclass · Accepted Answer

$x+3y+26=0$

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(D) The point $P$ on the curve $y=x^{2}+7x+2$ nearest to the line $y=3x-3$ is the point where the tangent to the curve is parallel to the given line.$1$. Find the slope of the tangent at any point $(x, y)$ on the curve:$\frac{dy}{dx} = \frac{d}{dx}(x^{2}+7x+2) = 2x+7$$2$. Since the tangent is parallel to the line $y=3x-3$,its slope must be equal to the slope of the line,which is $3$:$2x+7 = 3$$2x = -4$$x = -2$$3$. Find the $y$-coordinate of $P$ by substituting $x=-2$ into the curve equation:$y = (-2)^{2} + 7(-2) + 2 = 4 - 14 + 2 = -8$So,the point $P$ is $(-2, -8)$.$4$. The normal at $P$ is perpendicular to the tangent. The slope of the tangent is $3$,so the slope of the normal is $-\frac{1}{3}$.$5$. The equation of the normal at $P(-2, -8)$ with slope $m = -\frac{1}{3}$ is:$y - (-8) = -\frac{1}{3}(x - (-2))$$y + 8 = -\frac{1}{3}(x + 2)$$3(y + 8) = -(x + 2)$$3y + 24 = -x - 2$$x + 3y + 26 = 0$

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:

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