Question

Given the equation of the parabola

The focus of the parabola is:



(0, 9)

(-9, 0)

(0, -9)

Answer 1

https://clackamasweb.owschools.com/media/g_geo_ccss_2014/9/28.gif

Answer 2

@juliet8

Answer 3

\[
\large \text{Parabolas}\\
\begin{array}{r|c|c}
&\text{Standard}&\text{Vertex}\\
\hline
\text{Forms}&y=ax^2+bx+c &y=a(x-h)^2+k\\
\text{Discriminant}& b^2-4ac & -4ak \\
\text{Roots}&x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} & x=h\pm\sqrt{-\frac ka}\\
\text{Vertex}&\left(-\frac b{2a}, c-\frac{b^2}{4a}\right) &\left(h,k\right)\\
\text{Focus}&\left(-\frac b{2a}, c-\frac{b^2-1}{4a}\right) &\left(h,k+\frac{1}{4a}\right)\\
\text{Directrix}&y= c-\frac{b^2+1}{4a} & y=k-\frac{1}{4a}\\
\end{array}
\]

Answer 4

so its the second one?

Answer 5

Another equation for a parabola is \(4py=x^2\).

Answer 6

In this case the focus is at \((0,p)\).

Answer 7

It turns out that \(4p = 1/a\).

Answer 8

in our case \(4p = -36\).

Answer 9

Find \(p\).

Answer 10

(0,-9)?

Answer 11

Yes.

Answer 12

Thanks:)

Answer 13

so the directrix of the parabola is -9? @wio

Answer 14

Directrix is \(y=-p\).

Answer 15

so \(y=-(-9) = 9\)

Answer 16

thnx:)