Answer
b).
Work Step by Step
If the major axis of the parabola is parallel to the y-axis then it is in the form of $k(y-a)=(x-b)^2$, where $(b,a)$ is the vertex of the parabola and $k$ is a constant (positive if open up, negative if open down).
If the major axis of the parabola is parallel to the x-axis then it is in the form of $k(x-a)=(y-b)^2$, where $(a,b)$ is the vertex of the parabola and $k$ is a constant. (positive if open right, negative if open left)
Here the major axis is parallel to the y-axis, $k=\pm4$ in all cases, but here it is open up, hence $k=4$ and the vertex is in $(0,0)$.
Hence the equation: $4(y-0)=(x-0)^2\\4y=x^2$
Thus the answer is b).