A rectangle #OABC# is drawn so that #O# is the vertex of parabola #y=x^2# and #OA# and #OB# are two

The parabola #y=x^2# is symmetric w.r.t. #y#-axis. Further as #OABC# is a rectangle, its each angle is #90^@# i.e. #m/_AOB=90^@#.

As the sides #AO# and #BO# pass through origin at #O(0,0)#, their equation will be will be of type #y=mx# and #y=-1/mx# or #x+my=0#.

Now intersection of #y=mx# with parabola will be given by #mx=x^2# i.e. #x=m# and the corresponding point is #(m,m^2)#.

Similarly intersection of #x+my=0# gives #x+mx^2=0# i.e. #x=-1/m# and the corresponding point is #(-1/m,1/m^2)#.

As #A(m,m^2)# and #B(-1/m,1/m^2)#, we must have #C(m-1/m,m^2+1/m^2)# and as

#(m-1/m)^2=m^2+1/m^2-2#, desired equation is #y=x^2+2# and answer is (A).

Below is shown the graph and rectangle relating to #m=2#.

graph{(y-x^2)(2y+x)(y-2x)(8x-4y+5)(2x+4y-20)(y-x^2-2)=0 [-4.84, 5.16, -0.38, 4.62]}

I think that name of the rectangle should be #OACB# not #OABC#

Let the coordinates of

A #->(t_1,t_1^2)#
B #->(t_2,t_2^2)#
C #->(h,k)#
O #->(0,0)#

Gradient of OA #=t_1^2/t_1=t_1#
Gradient of OB #=t_2^2/t_2=t_2#

As OA and OB are adjacent sides of the rectangle then product of their gradients should be #-1#
Hence #t_1*t_2=-1... .(1)#

Now the diagonals ºC and #AB# should bisect each other.
So coordinates of mid point of #OC->(h/2,k/2)#

And coordinates of mid point of #AB->((t_1+t_2)/2,(t_1^2+t_2^2)/2)#

Hence #h=t_1+t_2 and k=(t_1^2+t_2^2)#

So #k=(t_1^2+t_2^2)#
#=>k=(t_1+t_2)^2-2t_1t_2#
#=>k=h^2-2(-1)#
#=>k=h^2+2#

Converting #(h,k)->(x,y)# we get the locus of #C# as follows

#color(green)(y=x^2+2#, which is option #(A)#