# sketch: x^2 = 8Y. indicate its vertex focus, directrix Parabola problem

### 1 Answer | Add Yours

`x^2=8y=>y=1/8x^2`

**Vertex**: x=-b/2a=>x=0, plug it in the original function, y=0. Hence **vertex (0,0)**

**Focus**:To find the p, we set 4p=coefecient of the unsquare part (use the given equation before we rewrote it) 4p=8 => p=2 => Focus is two units above the vertex => **Focus (0,2)**.

**Directrix**: is 2 units below the parabola =>** y=-2**