Unit 4, Lesson 19
Equations and Graphs
Integrated Math 2
19.1
Warm-up
Instructional Routines
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NoneActivity Narrative
Students find distances from points on a parabola to the parabola’s focus using the fact that this distance is equal to the distance between the point and the parabola’s directrix. This will be helpful in an upcoming activity in which students write an equation for a parabola.
Launch
NoneActivity
NoneThe image shows a parabola with focus and directrix (the -axis). Points , , and are on the parabola.
Parabola with focus at negative 2 comma 2 and directrix y equals 0, the x axis. Point A is 2 comma 5, point B is negative 8 comma 10, point C is negative x comma y and all points are on the parabola.
Without using the Pythagorean Theorem, find the distance from each plotted point to the parabola’s focus. Explain your reasoning.
Student Response
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Building on Student Thinking
If students are stuck, suggest that they refer to the definition of a parabola on their reference chart. Ask if there are any distances they can easily find without needing the Pythagorean Theorem.
Activity Synthesis
The goal is for students to summarize their observations about the general point . Display this image.
Ask students what we know about these two distances. (The distances are equal. Each distance is units.)
Lesson Synthesis
The goal of the discussion is for students to review how to find an equation of a parabola from the focus and directrix. Display this image of a parabola for all to see.
- “What theorem helps you find the distance between the focus at and any other point at ?” (the Pythagorean Theorem)
- "What is an equation that uses the Pythagorean Theorem to show the relationship between the focus at , a point at , and the distance between the two points?” ()
- “What is an equation that shows the distance between a point at and the directrix at ?” (If , then . If , then .)
- “How can you use the equations you found to write an equation for the parabola?” (Because the distance from a point to the focus must be the same as the distance to the directrix, I can set equal the expressions that are equal to . In this case, .)
- “How would you rewrite the equation into vertex form?” (I could expand the terms with in them and rearrange the equation to solve for .)
Student Lesson Summary
The parabola in the image consists of all the points that are the same distance from the point as they are from the line .
Suppose we want to write an equation for the parabola—that is, an equation that says a given point is on the curve. We can draw a right triangle whose hypotenuse is the distance between point and the focus, .
Parabola with a focus of 1 comma 4 and directrix y equals 0. All points on parabola are same distance from point 1 comma 4. The point x comma y is on the parabola. Triangle consists of sides labeled y for the hypotenuse, x minus 1 and y minus 4. Parabola contains point 1 comma 2.
The distance from to the directrix, or the line , is units. By definition, the distance from to the focus must be equal to the distance from the point to the directrix. So, the distance from to the focus can be labeled with . To find the lengths of the legs of the right triangle, subtract the corresponding coordinates of the point and the focus, .
Substitute the expressions for the side lengths into the Pythagorean Theorem to get an equation defining the parabola.
To get the equation looking more familiar, rewrite it in vertex form, or where is the vertex.