Eqn of normal at point ' t' of the parabola x^2 = 4by is
x + ty = 2bt + bt^3 .................( 1 )
Eqn of tangent at ( x1 , y1 ) of the parabola y^2 = 4ax is
yy1 = 2a ( x + x1 ) ................... ( 2 )
comparing ( 2 ) with ( 1 ) , we get
2a = - y1 / t = 2ax1 / ( 2bt + bt^3 ) ................ ( 3 )
Now( x1 , y1 ) satisfies y1^2 = 4ax1
so , from (3) we get
bt^2 - at + 2b = 0
we will get two normals provided discriminant of the above eqn > 0
giving a^2 > 8b^2 ( proved )