+4622 Ah, here it is!
The vertex has to be right in between them, so the vertex is \((0, 2)\). Now, we use the equation \((x-h)^2=4p(y-k)\). \(p = -2\) , because that is the directed distance between the vertex and focus. We get than \(x^2 = -8(y-2)\), so
\(\boxed{y = -\frac{1}{8}x^2+2}\) is the desired equation.
+2448 Thank you so much, but my teacher says it's \(y=- \frac{x^2}{8}+2\)
2 questions, plz help!
