Normals are drawn from the point P with slopes m1, m2, m3 to the parabola y2 = 4x. If locus of P with m1m2 = α is a part of parabola itself, then find α.

.3is y = mx – 2am – am = 4ax.2yThe equation of the normal to the parabola 

Hence, the equation of the normal to the parabola y2 = 4x is y = mx – 2m – m3.

Now, if it passes through (h, k), then we have k =  mh – 2m – m3

This can be written as m3 + m(2-h) + k = 0 …. (1)

Here, m1 + m2 + m3 = 0

m1m2 + m2m3 + m3m1 = 2 - h

Also, m1m2m3 = -k, where m1m2 = α

Now, this gives m3 = -k/α and this must satisfy equation (1).

Hence, we have (-k/α)3 + (-k/α)(2-h) + k = 0

Solving this, we get k2 = α2h – 2α2 + α3

And so we have y2 = α2x – 2α2 + α3

On comparing this equation with y2 = 4x we get

α2 = 4 and -2α2 + α3 = 0

This gives α = 2.