Question

# 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 $$4k - 2p^{2} =$$

A
1
B
0
C
1
D
2

Solution

## The correct option is D $$0$$$$y^{2} = 12x$$ has normal $$x + y = k$$$$y = tx + 2at + at^{3}$$Here, $$a = 3$$$$y = -tx + 6t + 3t^{3}$$ comparing$$\Rightarrow y = k - x$$So, $$t = 1$$$$\Rightarrow k= 6t + 3t^{3} = 9$$distance of focus from normal,$$P = \dfrac {|3(1) - 9|}{\sqrt {2}} = \dfrac {6}{\sqrt {2}}$$So, $$4k -2p^{2} = 36 - 2\left (\dfrac {36}{2}\right ) = 0$$Mathematics

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