Let x+y=k be a normal to the parabola y^2=12x | vedclass

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(B) The equation of the parabola is $y^2=12x$. Comparing with $y^2=4ax$,we get $a=3$.The equation of a normal to the parabola $y^2=4ax$ with slope $m$

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(B) The equation of the parabola is $y^2=12x$. Comparing with $y^2=4ax$,we get $a=3$.The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is $y=mx-2am-am^3$.Given the normal is $x+y=k$,which can be written as $y=-x+k$. Thus,$m=-1$.Substituting $m=-1$ and $a=3$ into the normal equation:$y = (-1)x - 2(3)(-1) - 3(-1)^3$$y = -x + 6 + 3$$y = -x + 9$Comparing $y = -x + 9$ with $x+y=k$,we get $k=9$.The focus of the parabola $y^2=12x$ is $S(a, 0) = (3, 0)$.The length of the perpendicular $p$ from the focus $(3, 0)$ to the line $x+y-9=0$ is:$p = \frac{|1(3) + 1(0) - 9|}{\sqrt{1^2 + 1^2}} = \frac{|-6|}{\sqrt{2}} = \frac{6}{\sqrt{2}}$.Therefore,$p^2 = \frac{36}{2} = 18$.Now,calculating $4k - 2p^2$:$4(9) - 2(18) = 36 - 36 = 0$.

Let $x+y=k$ be a normal to the parabola $y^2=12x$. If $p$ is the length of the perpendicular from the focus of the parabola onto this normal,then $4k-2p^2$ is equal to

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