The area of a square is represented by the equation \begin{align*}y = 9x^2\end{align*}

### Parabolas with Vertex at the Origin

You already know that the graph of a parabola has the parent graph \begin{align*}y=x^2\end{align*}**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 \begin{align*}y=ax^2\end{align*}

Notice, that when the parabola opens to the left or right, the \begin{align*}y\end{align*}

#### Solve the following problems

Analyze the equation \begin{align*}y^2=-12x\end{align*}

To find the focus and directrix, we need to find \begin{align*}p\end{align*}

\begin{align*}-12&=4p \\
-3&=p\end{align*}

Because \begin{align*}y\end{align*}

Notice that the points \begin{align*}(-3, 6)\end{align*}

The focus of a parabola is \begin{align*}\left(0, \frac{1}{2}\right)\end{align*}. Find the equation of the parabola.

Because the \begin{align*}p\end{align*} value is the \begin{align*}y\end{align*}-value and positive, this parabola is going to open up. So, the general equation is \begin{align*}x^2=4py\end{align*}. Plugging in \begin{align*}\frac{1}{2}\end{align*} for \begin{align*}p\end{align*}, we have \begin{align*}x^2=4 \cdot \frac{1}{2}y\end{align*} or \begin{align*}x^2=2y\end{align*}.

Find the equation of the parabola below.

The equation of the directrix is \begin{align*}y = 5\end{align*}, which means that \begin{align*}p = -5\end{align*} and the general equation will be \begin{align*}x^2=4py\end{align*}. Plugging in -5 for \begin{align*}p\end{align*}, we have \begin{align*}x^2=-20y\end{align*}.

### Examples

#### Example 1

Earlier, you were asked what is the focus and directrix of the equation.

To find the focus and directrix, we need to solve for \begin{align*}x^2\end{align*} and then find \begin{align*}p\end{align*}.

\begin{align*}y = 9x^2\\ \frac{1}{9} y = x^2\end{align*}

We can now set \begin{align*}\frac{1}{9} = 4p\end{align*} and solve for \begin{align*}p\end{align*}.

\begin{align*}\frac{1}{9} = 4p \\ \frac{1}{36}=p\end{align*}

Therefore, the focus is \begin{align*}(0, \frac{1}{36})\end{align*} and the directrix is \begin{align*}y = -\frac{1}{36}\end{align*}.

#### Example 2

Determine if the parabola \begin{align*}x^2=-2y\end{align*} opens up, down, left or right.

Down; \begin{align*}p\end{align*} is negative and \begin{align*}x\end{align*} is squared.

#### 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*}
- In the
*Quadratics*chapter, 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. 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*}.*Challenge*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).*Real World Application*

### Answers for Review Problems

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