The normal at a point on the parabola y^2=4x | vedclass

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(D) The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.For the parabola $y^2=4x$,we have $a=1$,so the equat

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$\sqrt{3}x-y-5\sqrt{3}=0$

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(D) The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.For the parabola $y^2=4x$,we have $a=1$,so the equation becomes $y=mx-2m-m^3$.Since the normal passes through $(5,0)$,we substitute $x=5$ and $y=0$:$0 = m(5) - 2m - m^3$$0 = 3m - m^3$$m(3 - m^2) = 0$Thus,the slopes are $m=0$ and $m=\pm\sqrt{3}$.For $m=\sqrt{3}$,the equation of the normal is $y=\sqrt{3}(x-5)$,which simplifies to $\sqrt{3}x-y-5\sqrt{3}=0$.For $m=-\sqrt{3}$,the equation of the normal is $y=-\sqrt{3}(x-5)$,which simplifies to $\sqrt{3}x+y-5\sqrt{3}=0$.For $m=0$,the equation of the normal is $y=0$.

The normal at a point on the parabola $y^2=4x$ passes through $(5,0)$. If there are two more normals to this parabola passing through $(5,0)$,then the equation of one of these normals is

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