Your homework assignment is to find the focus of the parabola

### Parabolas with Vertex at (h, k)

You learned in the *Quadratic Functions* chapter that parabolas don’t always have their vertex at

Recall from the previous concept that the equation of a parabola is *Quadratic Functions* chapter, we learned that the vertex form of a parabola is

If the parabola is horizontal, then the equation will be

Finding the focus and directrix are a little more complicated. Use the extended table (from the previous concept) below to help you find these values.

Notice that the way we find the focus and directrix does not change whether

#### Solve the following problems

Analyze the equation

First, because *right* because 8 is positive, meaning that

Graph the parabola from Example A. Plot the vertex, axis of symmetry, focus, and directrix.

First, plot all the critical values we found from Example A. Then, determine a set of symmetrical points that are on the parabola to make sure your curve is correct. If

It is important to note that parabolas with a horizontal orientation are not functions because they do not pass the vertical line test.

The vertex of a parabola is

First, let’s determine the orientation of this parabola. Because the directrix is horizontal, we know that the parabola will open up or down (see table/pictures above). We also know that the directrix is *above* the vertex, making the parabola open down and

To find

Now, using the general form,

### Examples

#### Example 1

Earlier, you were asked who is correct.

This parabola is of the form *h*, *k*, and *p*.

If we compare

1.

2. \begin{align*}-12 = 4p\end{align*} or \begin{align*}p = -3\end{align*}

3. \begin{align*}5 = k\end{align*}

From these facts we can find \begin{align*}k + p = 5 + (-3) = 2\end{align*}.

Therefore, the focus of the parabola is \begin{align*}(-4, 2)\end{align*} and Carlos is correct.

#### Example 2

Find the vertex, focus, axis of symmetry and directrix of \begin{align*}(x+5)^2=2(y+2)\end{align*}.

The vertex is \begin{align*}(-5, -2)\end{align*} and the parabola opens up because \begin{align*}p\end{align*} is positive and \begin{align*}x\end{align*} is squared. \begin{align*}4p = 2\end{align*}, making \begin{align*}p = 2\end{align*}. The focus is \begin{align*}(-5, -2 + 2)\end{align*} or \begin{align*}(-5, 0)\end{align*}, the axis of symmetry is \begin{align*}x = -5\end{align*}, and the directrix is \begin{align*}y =-2 - 2\end{align*} or \begin{align*}y = -4\end{align*}.

#### Example 3

Graph the parabola from #1.

#### Example 4

Find the equation of the parabola with vertex \begin{align*}(-5, -1)\end{align*} and focus \begin{align*}(-8, -1)\end{align*}.

The vertex is \begin{align*}(-5, -1)\end{align*}, so \begin{align*}h = -5\end{align*} and \begin{align*}k = -1\end{align*}. The focus is \begin{align*}(-8, -1)\end{align*}, meaning that that parabola will be horizontal. We know this because the \begin{align*}y\end{align*}-values of the vertex and focus are both -1. Therefore, \begin{align*}p\end{align*} is added or subtracted to \begin{align*}h\end{align*}.

\begin{align*}(h+p,k) \rightarrow (-8,-1)\end{align*} we can infer that \begin{align*}h+p=-8 \rightarrow -5+p=-8\end{align*} and \begin{align*}p=-3\end{align*}

Therefore, the equation is \begin{align*}(y-(-1))^2=4(-3)(x-(-5)) \rightarrow (y+1)^2=-12(x+5)\end{align*}.

### Review

Find the vertex, focus, axis of symmetry, and directrix of the parabolas below.

- \begin{align*}(x+1)^2=-3(y-6)\end{align*}
- \begin{align*}(x-3)^2=y-7\end{align*}
- \begin{align*}(y+2)^2=8(x+1)\end{align*}
- \begin{align*}y^2=-10(x-3)\end{align*}
- \begin{align*}(x+6)^2=4(y+8)\end{align*}
- \begin{align*}(y-5)^2=- \frac{1}{2}x\end{align*}
- Graph the parabola from #1.
- Graph the parabola from #2.
- Graph the parabola from #4.
- Graph the parabola from #5.

Find the equation of the parabola given the vertex and either the focus or directrix.

- vertex: \begin{align*}(2, -1)\end{align*}, focus: \begin{align*}(2, -4)\end{align*}
- vertex: \begin{align*}(-3, 6)\end{align*}, directrix: \begin{align*}x = 2\end{align*}
- vertex: \begin{align*}(6, 10)\end{align*}, directrix: \begin{align*}y = 9.5\end{align*}
focus: \begin{align*}(-1, -2)\end{align*}, directrix: \begin{align*}x = 3\end{align*}*Challenge*Rewrite the equation of the parabola, \begin{align*}x^2-8x+2y+22=0\end{align*}, in standard form by completing the square. Then, find the vertex. (For a review, see the*Extension**Completing the Square When the Leading Coefficient Equals 1*concept.)

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.2.