The correct option is D T∈[−∞,−2√2]∪[2√2,∞]
Given parabola is y2=4ax
Equation of normal at (at2,2at) is
y+tx=2at+at3 ⋯(1)
This cuts again at (aT2,2aT), so
2aT+taT2=2at+at3⇒2T+tT2=2t+t3 (∵a≠0)⇒t(t2−T2)+2(t−T)=0⇒(t−T)(t(t+T)+2)=0⇒t2+Tt+2=0 (∵t≠T)
As t∈R, so
D≥0⇒T2−8≥0∴T∈[−∞,−2√2]∪[2√2,∞]
Alternate solution :
If normal at (at2,2at) again cuts the parabola at (aT2,2aT), then
T=−t−2tt2+Tt+2=0
As t∈R, so
D≥0⇒T2−8≥0∴T∈[−∞,−2√2]∪[2√2,∞]