The correct option is D y2=4a(x−8a)
Given parabola is y2=4ax
Let the vertex be O=(0,0) and the chords be OA and OB
Where A=(at21,2at1),B=(at22,2at2)
Now, the slope of the line OA is
m1=2t1
Slope of the line OB is
m2=2t2⇒ m1.m2=−1 (∵ OA⊥OB)⇒t1t2=−4
Let the coordinates of the vertex opposite to the vertex of parabola be C=(h,k)
As the diagonals of a rectangle bisect each other, so
(h2,k2)=(a(t21+t22)2,2a(t1+t2)2)⇒ha=t21+16t21, k2a=t1−4t1⇒(k2a)2=ha−8
Hence, the locus is
y2=4a(x−8a)