Parabolic Psycho DK again

true
also remember that if three normals can be drawn from a pt. (h,k) to the parabola y²=4ax then 27ak²<4(h-2a)³

let P(h,k) be any pt. and y²=4ax be the given parabola.
the eqn of any normal to the parabola is y=mx-2am-am³
=>am³+m(2a-h) +k =0
this is a cubic and has three roots(say,m₁,m₂,m₃) !if the three normals are real, then the three roots are also real.
thus m₁²+m₂²+m₃² >0 (for any values of m₁,m₂,m₃)
=>(m₁+m₂+m₃)³ - 2>0
=>(0)² - 2/a * (2a-h) >0
=>h>2a

Great .... Thanks

one of the roots of am³ +m(2a-h)=0 is obviously zero. (the k is zero, if the point is on the axis)
so we get
am² = h-2a
Obviously, h > 2a since m² is positive for real m.