Answer
$b$
Work Step by Step
1. If the major axis of the parabola is parallel to the $y$-axis then it has the form of $(x=h)^2=4a(y-k)$.
where $(h, k)$ is the vertex of the parabola and $4a$ is a constant which will be positive if the graph opens up and negative if it opens down.
2. If the major axis of the parabola is parallel to the x-axis, then it has the form of $4a(x-h)=(y-k)^2$, where $(h, k)$ is the vertex of the parabola and $4a$ is a constant which will be positive if the graph opens right and negative if it opens left.
Since the major axis is parallel to the $y$-axis and opens up, in all cases $4a=\pm4$ and the vertex is: $(0,0)$.
Thus, the equation becomes : $4(y-0)=(x-0)^2 \implies x^2=4y$
So, our answer is $\bf{b}$.
