The correct option is C infinite such parabolas exist
From y=2x x=1,⇒y=2
Now putting (1,2) in the parabola equation
2=a+b+c⋯(1)
Equation of tangent at (x1,y1) for given parabola is
y+y12=axx1+b(x+x1)2+c
putting (x1,y1)≡(1,2), we get
y=(2a+b)x+(b+2c−2)
Comparing with y=2x
2a+b=2...(2)
b+2c−2=0⋯(3)
Solve (1),(2),(3), we get
c=a,b=2(1−a)
Hence the parabola will become
y=ax2+2(1−a)x+a
Now from here for different values of a except 0 we will have different parabolas which have tangent as y=2x