Hello, maxboy0801!
Learn this . . .
The general form is:
. \(\displaystyle x^2\,=\,4py\)
. . where \(\displaystyle p\) is the <u>directed</u> distance from the vertex to the focus.
Example:
.\(\displaystyle y\,=\,\frac{1}{8}x^2\)
. . Rewrite:
.\(\displaystyle x^2\,=\,8y\)
. . \(\displaystyle 4p\) is the coefficient of \(\displaystyle y\), so we have:
.\(\displaystyle 4p\,=\,8\;\Rightarrow\;p\,=\,+2\)
. . So we go <u>up</u> 2 units to find the focus: \(\displaystyle (0,\,2)\)
Example:
.\(\displaystyle y\,=\,-2x^2\)
. . Rewrite:
.\(\displaystyle x^2\,=\,-\frac{1}{2}y\)
. . We have:
.\(\displaystyle 4p\,=\,-\frac{1}{2}\;\;\Rightarrow\;\;p\,=\,-\frac{1}{8}\)
. . So we go <u>down</u> \(\displaystyle \frac{1}{8}\) unit to the focus: \(\displaystyle (0,\,-\frac{1}{8})\)
The focus of the parabola \(\displaystyle y\,=\,x^2\) is__.
You should be able to handle this now . . .
Rewrite:
.\(\displaystyle x^2\,=\,y\) . . . now find
p.
