The equation of the normal to the parabola y^ | vedclass

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(A) The given equation of the parabola is $y^2=4x$,which implies $a=1$.The equation of the normal to the parabola $y^2=4ax$ with slope $m$ is given by

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$3x-y=33$

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(A) The given equation of the parabola is $y^2=4x$,which implies $a=1$.The equation of the normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.Substituting $a=1$,the equation of the normal is $y=mx-2m-m^3 \dots(I)$.The given line is $x+3y+1=0$,which can be written as $y=-\frac{1}{3}x-\frac{1}{3}$. The slope of this line is $m_1 = -\frac{1}{3}$.Since the normal is perpendicular to this line,the product of their slopes must be $-1$. Let $m$ be the slope of the normal,then $m 	imes (-\frac{1}{3}) = -1$,which gives $m=3$.Substituting $m=3$ into equation $(I)$:$y = 3x - 2(3) - (3)^3$$y = 3x - 6 - 27$$y = 3x - 33$$3x - y = 33$.

The equation of the normal to the parabola $y^2=4x$ which is perpendicular to the line $x+3y+1=0$ is

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