The equation of parabola be y2=4ax
let the point P be (at2,2at)
PN is ordinate ⇒N(at2,0)
Equation of straight line bisecting NP is
y=at
substituting y in equation of parabola
a2t2=4ax⇒x=at24
So the coordinates of Q are (at24,at)
Equation of NQ is
y−0=at−0at24−at2(x−at2)y=−43t(x−at2)
Put x=0
y=−43t(0−at2)⇒y=4at3
⇒AT=4at3NP=2atATNP=4at32at=23AT=23NP