The correct option is C x2=4by
Let the three tangent equations to parabola y2=4ax are
yt1=x+at21
yt2=x+at22
yt3=x+at23
Intersection points of these tangents are A(at1t2,a(t1+t2)) , B(at2t3,a(t2+t3)) , C(at1t3,a(t1+t3))
Let A and B lie on x2=4by
a2t21t22=4ab(t1+t2)⋯(1)
a2t22t23=4ab(t2+t3)⋯(2)
Now eliminating t3 from above equations and hence bringing relationship between t1,t2
From (1)÷(2), we get
t21t23=t1+t2t2+t3
⇒t2(t21−t23)=t1t3(t3−t1)
⇒t2=−t1t3(t1+t3)⋯(3)
From (1)−(2)
a2t22(t21−t23)=4ab(t1−t3)
a2[−t1t3(t1+t3)]2(t1+t3)=4ab⇒(at1t3)2=4b[a(t1+t3)]
Hence C(at1t3,a(t1+t3)) will also lie on x2=4by