When working with parabolas in the past you probably used vertex form and analyzed the graph by finding its roots and intercepts. There is another way of defining a parabola that turns out to be more useful in the real world. One of the many uses of parabolic shapes in the real world is satellite dishes. In these shapes it is vital to know where the receptor point should be placed so that it can absorb all the signals being reflected from the dish.

Where should the receptor be located on a satellite dish that is four feet wide and nine inches deep?

#### Watch This

http://www.youtube.com/watch?v=k7wSPisQQYs James Sousa: Conic Sections: The Parabola part 1 of 2

http://www.youtube.com/watch?v=CKepZr52G6Y James Sousa: Conic Sections: The Parabola part 2 of 2

#### Guidance

The definition of a parabola is the collection of points equidistant from a point called the focus and a line called the directrix.

Notice how the three points \begin{align*}P_1, P_2, P_3\end{align*}

\begin{align*}\overline{FP_1} &= \overline{P_1Q_1}\\
\overline{FP_2} &= \overline{P_2Q_2}\\
\overline{FP_3} &= \overline{P_3Q_3}\end{align*}

There are two graphing equations for parabolas that will be used in this concept. The only difference is one equation graphs parabolas opening vertically and one equation graphs parabolas opening horizontally. You can recognize the parabolas opening vertically because they have an \begin{align*}x^2\end{align*}

Note that the vertex is still \begin{align*}(h, k)\end{align*}

Once you put the parabola into this graphing form you can sketch the parabola by plotting the vertex, identifying \begin{align*}p\end{align*}

**Example A**

Identify the following conic, put it into graphing form and identify its vertex, focal length \begin{align*}(p)\end{align*}

\begin{align*}2x^2+16x+y=0\end{align*}

**Solution: ** This is a parabola because the \begin{align*}y^2\end{align*}

\begin{align*}x^2+8x &= -\frac{1}{2}y\\
x^2+8x+16 &= -\frac{1}{2}y+16\\
(x+4)^2 &= -\frac{1}{2}(y-32)\\
(x+4)^2 &= -4 \cdot \frac{1}{8} (y-32)\end{align*}

The vertex is (-4, 32). The focal length is \begin{align*}p=\frac{1}{8}\end{align*}

**Example B**

Sketch the following parabola and identify the important pieces of information.

\begin{align*}(y+1)^2=4 \cdot \frac{1}{2} \cdot (x+3)\end{align*}

**Solution: **

The vertex is at (-3, -1). The parabola is sideways because there is a \begin{align*}y^2\end{align*}

**Example C**

What is the equation of a parabola that has a focus at (4, 3) and a directrix of \begin{align*}y=-1\end{align*}

**Solution: **It would probably be useful to graph the information that you have in order to reason about where the vertex is.

The vertex must be halfway between the focus and the directrix. This places it at (4, 1). The focal length is 2. The parabola opens upwards. This is all the information you need to create the equation.

\begin{align*}(x-4)^2 = 4 \cdot 2 \cdot (y-1)
\end{align*}

OR \begin{align*}(x-4)^2 = 8(y-1)\end{align*}

**Concept Problem Revisited**

Where should the receptor be located on a satellite dish that is four feet wide and nine inches deep?

Since real world problems do not come with a predetermined coordinate system, you can choose to make the vertex of the parabola at (0, 0). Then, if everything is done in inches, another point on the parabola will be (24, 9). *(Many people might mistakenly believe the point* (48, 9) *is on the parabola but remember that half this width stretches to* (-24, 9) *as well.)* Using these two points, the focal width can be found.

*\begin{align*}(x-0)^2 &= 4p(y-0)\\
(24-0)^2 &= 4p(9-0)\\
\frac{24^2}{4 \cdot 9} &= p\\
16 &= p\end{align*} (x−0)2(24−0)22424⋅916=4p(y−0)=4p(9−0)=p=p*

The receptor should be sixteen inches away from the vertex of the parabolic dish.

#### Vocabulary

The ** focus** of a parabola is the point that the parabola seems to curve around.

The ** directrix** of a parabola is the line that the parabola seems to curve away from.

A ** parabola** is the collection of points that are equidistant from a fixed focus and directrix.

#### Guided Practice

1. What is the equation of a parabola with focus at (2, 3) and directrix at \begin{align*}y=5\end{align*}

2. What is the equation of a parabola that opens to the right with focal width from (6, -7) to (6, 12)?

3. Sketch the following conic by putting it into graphing form and identifying important information.

\begin{align*}y^2-4y+12x-32=0\end{align*}

**Answers:**

1. The vertex must lie directly between the focus and the directrix, so it must be at (2, 4). The focal length is therefore equal to 1. The parabola opens downwards.

\begin{align*}(x-2)^2=-4 \cdot 1 \cdot (y-4)\end{align*}

2. The focus is in the middle of the focal width. The focus is \begin{align*}\left(6, \frac{5}{2}\right)\end{align*}

\begin{align*}\left(y-\frac{5}{2}\right)^2=4 \cdot \frac{19}{4} \cdot \left(x-6+\frac{19}{4}\right)\end{align*}

3. \begin{align*}y^2-4y+12x-32 = 0
\end{align*}

\begin{align*}y^2-4y &= -12x+32\\
y^2-4y+4 &= -12x+32+4\\
(y-2)^2 &= -12(x-3)\\
(y-2)^2 &= -4 \cdot 3 \cdot (x-3)\end{align*}

The vertex is at (3, 2). The focus is at (0, 2). The directrix is at \begin{align*}x=6\end{align*}.