The equation of the tangent to the parabola y | vedclass

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(D) The given line is $y = 3x + 7$,which has a slope $m_1 = 3$.Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes m_

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$3y + x + 36 = 0$

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(D) The given line is $y = 3x + 7$,which has a slope $m_1 = 3$.Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes m_1 = -1$,so $m = -\frac{1}{3}$.The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Comparing $y^2 = 16x$ with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.Substituting $a = 4$ and $m = -\frac{1}{3}$ into the tangent equation:$y = -\frac{1}{3}x + \frac{4}{-1/3}$$y = -\frac{1}{3}x - 12$Multiplying by $3$ gives $3y = -x - 36$,which simplifies to $x + 3y + 36 = 0$.

The equation of the tangent to the parabola $y^2 = 16x$,which is perpendicular to the line $y = 3x + 7$,is

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