Prove that the locus of the point of intersection of the normals at the ends of a system of parallel cords of a parabola is a straight line which is a normal to the curve.

Explanation

Consider chord PQ joining points `P(at_(1)^(2),2at_(1))andQ(at_(2)^(2),2at_(2))` on parabola `y^(2)=4ax`.
Slope of PQ is m `=(2a(t_(2)-t_(1)))/(a(t_(2)^(2)-t_(1)^(2)))=(2)/(t_(2)+t_(1))`
Point of intersection of normals at points P and Q is
`R(2a+a(t_(1)^(2)+t_(2)^(2)+t_(1)t_(2)),at_(1)t_(2)(t_(1)+t_(2)))-=(h,k)`
`:." "h-2a=a((t_(1)+t_(2))^(2)-t_(1)t_(2))andk=-at_(1)t_(2)(t_(1)+t_(2))`
Putting `t_(1)+t_(2)=(2)/(m)`, we get
`h-2a=(4a)/(m^(2))-at_(1)t_(2)andk=-(2a)/(m)t_(1)t_(2)`
Eliminating `t_(1)t_(2)`, we get
`h-2a=(4a)/(m^(2))+(mk)/(2)`
So, locus of R is
`x-2a=(4a)/(m^(2))+(m)/(2)y`
`or" "y=(2)/(m)x-(4a)/(m)-(8a)/(m^(3))`
`or" "y=m'x-2am'-am'^(3)," where "m'=(2)/(m)`
Thus, locus of point R is normal to the parabola.