The line x + y = 6 is a normal to the parabol | vedclass

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(C) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$.The equation of the line is $x + y = 6$,whi

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$(2, 4)$

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(C) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$.The equation of the line is $x + y = 6$,which can be written as $y = -x + 6$. Comparing this with $y = mx + c$,we get $m = -1$.The condition for the line $y = mx + c$ to be a normal to the parabola $y^2 = 4ax$ is $c = -2am - am^3$.Substituting the values: $c = -2(2)(-1) - 2(-1)^3 = 4 + 2 = 6$. This confirms the line is a normal.The point of contact for a normal with slope $m$ is $(am^2, -2am)$.Substituting $a = 2$ and $m = -1$,we get the point $(2(-1)^2, -2(2)(-1)) = (2, 4)$.

The line $x + y = 6$ is a normal to the parabola $y^2 = 8x$ at the point

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