The correct option is A y(dydx)2+2xdydx−y=0
∵ focus and vertex both lie on the x−axis
∴ axis of the parabola will be x−axis
Let coordinates of the vertex is (−a,0)
Equation of the Directrix will be x+2a=0
Let any point on the parabola be P(x,y)
Now PS=PM
PS2=PM2x2+y2=(x+2a)2x2+y2=x2+4a2+4axy2=4a(a+x)
Differentiating with respect to x
2ydydx=4aa=y2⋅dydx∴y2=2y⋅dydx(y2⋅dydx+x)y(dydx)2+2xdydx−y=0