# Lesson 8

Equations and Graphs

• Let’s write an equation for a parabola.

### 8.1: Focus on Distance

The image shows a parabola with focus $$(\text-2,2)$$ and directrix $$y=0$$ (the $$x$$-axis). Points $$A$$, $$B$$, and $$C$$ are on the parabola.

Without using the Pythagorean Theorem, find the distance from each plotted point to the parabola’s focus. Explain your reasoning.

### 8.2: Building an Equation for a Parabola

The image shows a parabola with focus $$(3,2)$$ and directrix $$y=0$$ (the $$x$$-axis).

1. Write an equation that would allow you to test whether a particular point $$(x,y)$$ is on the parabola.
2. The equation you wrote defines the parabola, but it’s not in a very easy-to-read form. Rewrite the equation to be in vertex form: $$y=a(x-h)^2+k$$, where $$(h,k)$$ is the vertex.

### 8.3: Card Sort: Parabolas

Your teacher will give you a set of cards with graphs and equations of parabolas. Match each graph with the equation that represents it.

In this section, you have examined points that are equidistant from a given point and a given line. Now consider a set of points that are half as far from a point as they are from a line.

1. Write an equation that describes the set of all points that are $$\frac{1}{2}$$ as far from the point $$(5,3)$$ as they are from the $$x$$-axis.
2. Use technology to graph your equation. Sketch the graph and describe what it looks like.

### Summary

The parabola in the image consists of all the points that are the same distance from the point $$(1,4)$$ as they are from the line $$y=0$$. Suppose we want to write an equation for the parabola—that is, an equation that says a given point $$(x,y)$$ is on the curve. We can draw a right triangle whose hypotenuse is the distance between point $$(x,y)$$ and the focus, $$(1,4)$$.

The distance from $$(x,y)$$ to the directrix, or the line $$y=0$$, is $$y$$ units. By definition, the distance from $$(x,y)$$ to the focus must be equal to the distance from the point to the directrix. So, the distance from $$(x,y)$$ to the focus can be labeled with $$y$$. To find the lengths of the legs of the right triangle, subtract the corresponding coordinates of the point $$(x,y)$$ and the focus, $$(1,4)$$. Substitute the expressions for the side lengths into the Pythagorean Theorem to get an equation defining the parabola.

$$(x-1)^2+(y-4)^2=y^2$$

To get the equation looking more familiar, rewrite it in vertex form, or $$y=a(x-h)^2+k$$ where $$(h,k)$$ is the vertex.

$$(x-1)^2+(y-4)^2=y^2$$

$$(x-1)^2+y^2-8y+16=y^2$$

$$(x-1)^2-8y+16=0$$

$$\text-8y=\text-(x-1)^2-16$$

$$y=\frac18 (x-1)^2+2$$

### Glossary Entries

• directrix

The line that, together with a point called the focus, defines a parabola, which is the set of points equidistant from the focus and directrix.