in this problem, we need to find the equation of a parabola given the vertex and (0,0) and the focus at (0,-1/11).
First, let's look at the standard form a parabola. This equation can sometimes vary depending on the curriculum you use.
[tex]\begin{gathered} (x-h)^2=4p(y-k)\text{ for a parabola with a vertical axis.} \\ \\ (y-k)^2=4p(x-h)\text{ for a parabola with a horizontal axis.} \end{gathered}[/tex]
This is what it looks like when you see the different parts of a parabola on a graph:
Notice the focus lines up with the vertex, so this parabola has a vertical axis. Since our Focus has an x-value of 0 and a y-value of -1/11, we are working with a vertical parabola that opens down.
The focus is always going to be inside the parabola.
Since this parabola opens down, and it has a vertex at (0,0), we can use a modified equation:
[tex]x^2=4py[/tex]
From the equation, the "p" value represents the distance from the vertex to the directrix and the focus.
In this case:
[tex]p=|\frac{-1}{11}|=\frac{1}{11}[/tex]
So now we have the equation
[tex]\begin{gathered} x^2=4(\frac{1}{11})y \\ \\ x^2=\frac{4}{11}y \end{gathered}[/tex]
And since the parabola opens down, we simply include a negative with our equation:
[tex]x^2=-\frac{4}{11}y[/tex]
