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Normals are drawn at points A, B, and C ...

Normals are drawn at points `A, B, and C` on the parabola `y^2 = 4x` which intersect at P. The locus of the point P if the slope of the line joining the feet of two of them is 2, is

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The equation of normal at `(r^(2),2t)` is
`y=-tx+2t+t^(3)`
If it passes through P(h,k), then
`t^(3)+t(2-h)-k=0` (1)
This equation has roots `t_(1),t_(2)andt_(3)` which are parameters of three feet of normal has three feet of normals on the parabola.
From equation (1), we have
`t_(1)+t_(2)+t_(3)=0` (2)
Given that slope of the line joining the feet of two the normals (say `A(t_(1))andB(t_(2))` is 2.
`:." "(2)/(t_(1)+t_(2))=2`
`rArr" "t_(1)+t_(2)=1`
So, from (2), we have `t_(3)=-1`, which is one of the roots of equation (1).
So, from (1) we have
-1+h-2-k=0
or x-y-3=0, which is required locus.
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