Answer:
The equation of the parabola is y=(1/8)x^2
![y=\dfrac{1}{8}x^2](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B1%7D%7B8%7Dx%5E2)
Step-by-step explanation:
We know the vertex (0,0) and the focus (0,2) of the parabola.
The equation in vertex form is written as:
![y=a(x-h)^2+k\\\\\text{vertex}=(h,k)](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk%5C%5C%5C%5C%5Ctext%7Bvertex%7D%3D%28h%2Ck%29)
Then, in this case, we have the equation:
![y=a(x-0)^2+0=ax^2](https://tex.z-dn.net/?f=y%3Da%28x-0%29%5E2%2B0%3Dax%5E2)
As the focus is (0,2), it is at a distance of 2 units from the vertex (0,0).
For the focus, we have the following equation:
![4p(y-k)=(x-h)^2](https://tex.z-dn.net/?f=4p%28y-k%29%3D%28x-h%29%5E2)
where p is the distance from the focus to the vertex (in this case, p=2).
h and k are the vertex coordinates, both 0.
So we have:
![4p(y-k)=(x-h)^2\\\\4(2)(y-0)=(x-0)^2\\\\8y=x^2\\\\y=\dfrac{1}{8}x^2](https://tex.z-dn.net/?f=4p%28y-k%29%3D%28x-h%29%5E2%5C%5C%5C%5C4%282%29%28y-0%29%3D%28x-0%29%5E2%5C%5C%5C%5C8y%3Dx%5E2%5C%5C%5C%5Cy%3D%5Cdfrac%7B1%7D%7B8%7Dx%5E2)