The correct option is D a(t+1t)2
If t1 is the other end of the focal chord, then the equation of the chord is y(t+t1)=2x+2att1
Since it passes through the focus (a, 0), we get 2a+2att1=0⇒tt1=−1⇒t1=−1t
Length of the chord = √(at2−at21)2+(2at−2at1)2=a√(t2−1t2)2+4(t+1t)2
=a(t+1t)√(t−1t)2+4=a(t+1t)2