# Lesson 8

Equations and Graphs

## 8.1: Focus on Distance (5 minutes)

### Warm-up

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.

### Student Facing

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.

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

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 $$(x,y)$$. Display this image for all to see.

Ask students what we know about these two distances. (The distances are equal. Each distance is $$y$$ units.)

## 8.2: Building an Equation for a Parabola (20 minutes)

### Activity

Students use the concepts they’ve explored in recent activities to write an equation for a parabola and rewrite it in vertex form. If many students struggle with the algebraic manipulation, consider working through some of the steps as a class.

### Launch

Representation: Representation: Internalize Comprehension. Use color coding and annotations to highlight connections between representations in a problem. For example, students can draw the equal distances to the focus and the directrix from $$(x,y)$$ or label the image to identify $$h$$ and $$k$$.
Supports accessibility for: Visual-spatial processing

### Student Facing

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.

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

If students struggle to write the initial equation, suggest they draw in a right triangle whose hypotenuse is the distance between $$(3,2)$$ and $$(x,y)$$, and write expressions for the lengths of the triangle’s legs. Then, suggest they look back to the warm-up for a way to label the hypotenuse.

If students don’t know how to start rewriting the equation into vertex form, suggest they begin by squaring the expression $$(y-2)^2$$. Then rearrange the terms so $$y$$ is alone on one side of the equation.

### Activity Synthesis

Display the two forms of the equation. If students ask about the name of the original form, tell them that there is no particular name for that form. It is simply a starting point.

$$(x-3)^2 + (y-2)^2 = y^2$$

$$y=\frac 14 (x-3)^2+1$$

Here are some questions for discussion:

• “There is an $$(x-3)^2$$ in each equation. What does this mean in each one?” (In the first equation, the 3 came from the $$x$$-coordinate of the focus. In the second equation, it’s the $$x$$-coordinate of the vertex.)
• “Was it just coincidence that the $$x$$-coordinate of the focus and vertex were the same? That is, if we tried another example, could we find one in which the $$x$$-coordinate is different?” (If the directrix is a horizontal line, then the vertex and the focus will always be on the same vertical of line—the axis of symmetry.)

Tell students that it’s possible to create parabolas using directrices that are vertical or even slanted. A vertical directrix is relatively easy to work with, but for slanted directrices, we need to use more complicated trigonometry than what we’ve yet learned. Explain that because of these complications, we won’t create a general form for the equation of a parabola like we did for a circle—we will just create equations for specific parabolas, as we did in this activity.

## 8.3: Card Sort: Parabolas (10 minutes)

### Activity

A sorting activity gives students opportunities to analyze representations, statements, and structures closely and make connections (MP2, MP7).

As students work, encourage them to refine their descriptions of the parabolas using more precise language and mathematical terms (MP6).

### Launch

Arrange students in groups of 2. Distribute one set of cards to each group. Give students time to work with their partner, followed by a whole-class discussion.

Conversing: MLR2 Collect and Display. As students work on this activity, listen for and collect the language students use to justify why they matched a graph of a parabola with an equation. Write the students’ words and phrases on a visual display. As students review the visual display, create bridges between current student language and new terminology. For example, the word “point” must be specified with the term “focus,” “vertex,” or “point on the parabola $$(x,y)$$.” This will help students use the mathematical language necessary to precisely describe the relationship between the focus, vertex, and the equation of the parabola.
Design Principle(s): Optimize output (for comparison); Maximize meta-awareness

### Student Facing

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

### Student Response

For access, consult one of our IM Certified Partners.

### Student Facing

#### Are you ready for more?

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.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

Select groups to share how they sorted their cards. Attend to the language that students use to describe the graphs and equations, giving them opportunities to describe the parabolas more precisely. Highlight the use of terms like focus, directrix, and distance.

## Lesson Synthesis

### Lesson Synthesis

The goal of the discussion is to summarize concepts from recent lessons. Ask students how the Pythagorean Theorem is related to circles and parabolas. (Both circles and parabolas are defined by distances. The Pythagorean Theorem allows us to calculate distances in the coordinate plane, so we can use it to write equations for circles and parabolas.)

Then, display these images for all to see. Ask students to write equations for each figure, then compare and contrast the results.

Sample responses:

• The equation for the circle is $$(x-6)^2+(y-3)^2=2^2$$. An equation for the parabola is $$(x-6)^2+(y-3)^2=(y-1)^2$$.
• The left sides of the equations are identical. Each represents the square of the distance from a point $$(x,y)$$ on the figure to the point $$(6,3)$$. In the circle, this point is the center. For the parabola, $$(6,3)$$ is the focus.
• The right sides of the equations are different. In the circle, the distance between $$(x,y)$$ and $$(6,3)$$ is the radius, so the right side of the circle’s equation is the square of the radius, or $$2^2$$. For the parabola, the distance between the point $$(x,y)$$ and $$(6,3)$$ must be the same as the distance between $$(x,y)$$ and the parabola’s directrix. The directrix is the line $$y=1$$, so the distance is $$(y-1)$$.

## 8.4: Cool-down - One More Equation (5 minutes)

### Cool-Down

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## Student Lesson Summary

### Student Facing

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.

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

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

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

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

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