If t1 and t2 are end of a focal chord of the parabola (y2=4ax) then t1.t2=−1
True
Given parabola B,
y2=4ax
focus=(a,0)
ends of focal chords A≡(at21,2at1)
B≡(at21,2at2)
AB is given by the equation,
2x−(t1+t2)y+2at1t2=0
This passes throug (a,0) since chord is focal chort,
2a−0+2at1t2=0
i.e., 2at1t2=0
∴ t1t2=−1
A natural corollary of this is that if one points is (at2,2at)
then other will be (at2,−2at)