The area of a square is represented by the equation

### Parabolas with Vertex at the Origin

You already know that the graph of a parabola has the parent graph **focus**, and a line called the **directrix**.

The focus is on the axis of symmetry and the vertex is halfway between it and the directrix. The directrix is perpendicular to the axis of symmetry.

Until now, we have been used to seeing the equation of a parabola like

Notice, that when the parabola opens to the left or right, the

Let's analyze the equation

To find the focus and directrix, we need to find

Because

Notice that the points

The focus of a parabola is

Because the

Now, let's find the equation of the parabola below.

The equation of the directrix is

### Examples

#### Example 1

Earlier, you were asked to find the focus and directrix of the equation

To find the focus and directrix, we need to solve for

We can now set

Therefore, the focus is

#### Example 2

Determine if the parabola

Down;

#### Example 3

Find the focus and directrix of \begin{align*}y^2=6x\end{align*}. Then, graph the parabola.

Solving for \begin{align*}p\end{align*}, we have \begin{align*}4p=6 \rightarrow p=\frac{3}{2}\end{align*}. Because \begin{align*}y\end{align*} is squared and \begin{align*}p\end{align*} is positive, the parabola will open to the right. The focus is \begin{align*}\left(\frac{3}{2},0\right)\end{align*} and the directrix is \begin{align*}x=- \frac{3}{2}\end{align*}.

#### Example 4

Find the equation of the parabola with directrix \begin{align*}x=- \frac{3}{8}\end{align*}.

If the directrix is negative and vertical \begin{align*}(x = )\end{align*}, we know that the equation is going to be \begin{align*}y^2=4px\end{align*} and the parabola will open to the right, making \begin{align*}p\end{align*} positive; \begin{align*}p=\frac{3}{8}\end{align*}. Therefore, the equation will be \begin{align*}y^2=4 \cdot \frac{3}{8} \cdot x \rightarrow y^2= \frac{3}{2}x\end{align*}.

### Review

Determine if the parabola opens to the left, right, up or down.

- \begin{align*}x^2=4y\end{align*}
- \begin{align*}y^2=- \frac{1}{2}x\end{align*}
- \begin{align*}x^2=-y\end{align*}

Find the focus and directrix of the following parabolas.

- \begin{align*}x^2=-2y\end{align*}
- \begin{align*}y^2=\frac{1}{4}x\end{align*}
- \begin{align*}y^2=-5x\end{align*}

Graph the following parabolas. Identify the focus and directrix as well.

- \begin{align*}x^2=8y\end{align*}
- \begin{align*}y^2=\frac{1}{2}x\end{align*}
- \begin{align*}x^2=-3y\end{align*}

Find the equation of the parabola given that the vertex is \begin{align*}(0, 0)\end{align*} and the focus or directrix.

- focus: \begin{align*}(4, 0)\end{align*}
- directrix: \begin{align*}x = 10\end{align*}
- focus: \begin{align*}\left(0, \frac{7}{2}\right)\end{align*}
- You have seen that earlier the basic parabolic equation was \begin{align*}y=ax^2\end{align*}. Now, we write \begin{align*}x^2=4py\end{align*}. Rewrite \begin{align*}p\end{align*} in terms of \begin{align*}a\end{align*} and determine how they affect each other.
**Challenge**Use the distance formula, \begin{align*}d=\sqrt{\left(x_2-x_1\right)^2- \left(y_2-y_1\right)^2}\end{align*}, to prove that the point \begin{align*}(4, 2)\end{align*} is on the parabola \begin{align*}x^2=8y\end{align*}.**Real World Application**A satellite dish is a 3-dimensional parabola used to retrieve sound, TV, or other waves. Assuming the vertex is \begin{align*}(0, 0)\end{align*}, where would the focus have to be on a satellite dish that is 4 feet wide and 9 inches deep? You may assume the parabola has a vertical orientation (opens up).

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.1.