The correct option is A x2a2+9y2b2=1
Given that : x2a2+y2b2=1
Let end points of double ordinates be P(α,β) and Q(α,−β)
Let the point which divides PQ in 1:2 is R(h,k)
Then, by internal section formula
h=α
And
k=1×β−2×β3 or k=2×β−1×β3⇒β=±3k
∵(α,β) lies on ellipse
∴h2a2+(±3k)2b2=1
Hence, the locus of R is
x2a2+9y2b2=1