The correct option is
A x22+y2=cGiven the equation of the parabola is
y=ax2......(1).
Now, differentiating both sides of (1) we get, dydx=2ax or, a=y′2x.
Using this in equation (1) we get,
y=xy′2......(2).
This is the differential equation corresponding to the parabola (1).
Now, the differential equation of the orthogonal trajectory of (1) is
y=−x2y′ [ Replacing y′ in (2) by −1y′]
or, 2y dy+xdx=0
Now integrating both sides we get,
y2+x22=c [ Where c is integrating constant]
This is the required equation of orthogonal trajectory to (1).