1 more focus question

The focus of the parabola y = x^2 is

would the facus be (0, 1/2) I really don't know what I am doing........

-Maxwell

A parabola consists of points equidistant from a focus and a directrix, a line. If the focus is at (0,b), the vertex at (0,0) and the directrix is at y=-b then when x = 2b the point on the parabola will be (2b,b)
y=x²
b=(2b)² = 4b²
divide by b
1=4b
b=1/4
so the focus is at (0,1/4)

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__.

Click to expand...

You should be able to handle this now . . .

Rewrite: .\(\displaystyle x^2\,=\,y\) . . . now find p.