Topics
12. Connecting Algebra and Geometry Through Coordinates
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United States of America
Grades 9-12

12.03 Parabolas in the coordinate plane

Lesson
Practice

Introduction

As we have seen in this chapter, Algebra and Geometry are topics that are closely related to one another. We will derive the equation of a parabola in this lesson using concepts we learned in  10.01 Distance and the coordinate plane  and connect it to concepts we learned in Algebra 1 lesson  10.03 Quadratic functions in vertex form  .

Definition of a parabola

A parabola is one of the special shapes you can make when you slice a cone on an angle. It is known as one of the conic sections.

Circle
Ellipse
Parabola
Hyperbola

Exploration

Move point A left and right.

  1. What do you notice?

  2. What do you wonder?

Loading interactive...

A parabola is an arch-shaped curve such that any point on the curve is equidistant from a fixed point (called the focus) and a fixed line (called the directrix).

Upward-facing parabola
Downward-facing parabola

We can determine if a point is on a parabola if we are given the coordinates of the focus and the equation of the directrix. We need to check that the distance from the point to the directrix and the distance from the point to the focus are equal.

Examples

Example 1

A parabola has a focus at \left(-4,1\right) and a directrix at y=-2. Determine if the following points are on the parabola:

a

\left(-4,-\dfrac{1}{2}\right)

Worked Solution
Create a strategy

Graphing the values we know, can help us visualize this situation.

-8
-7
-6
-5
-4
-3
-2
-1
x
-3
-2
-1
1
2
3
y

Looking at our graph, we can see that the point is directly between the focus and the directrix.

Apply the idea

Mathematically, we can determine the point is \left|1-\left(-\dfrac{1}{2}\right)\right|=\dfrac{3}{2} away from the directrix and \left|-\dfrac{1}{2}-\left(-2\right)\right|=\dfrac{3}{2} away from the focus.

Because the distance from the point to the focus and the distance from the point to the directrix are the same, this point does lie on the parabola.

Reflect and check

Because the directrix is in the form y=c, we know it will be a horizontal line. The focus lies above the line, so the parabola opens upward.

Whenever the point lies halfway between the focus and directrix and lies on the axis of symmetry, the point is actually the vertex of the parabola. Recall the axis of symmetry is the vertical line passing through the x-value of the focus and vertex.

We did not know the vertex in this problem, but we were given the focus. So, the line of symmetry is x=-4, and the x-value of our point is also -4. That means the point does lie on the axis of symmetry, and we found that it is halfway between the focus and directrix.

b

\left(4,10\right)

Worked Solution
Create a strategy

The graph of this situation looks like this:

-8
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-5
-4
-3
-2
-1
1
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x
-2
-1
1
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10
y

We can find the distance of the point to the directrix by using absolute values. We can find the distance of the point to the focus by using the distance formula.

Apply the idea

The distance from the point to the directrix is

d_1=\left|10-\left(-2\right)\right|=12

The distance from the point to the focus is

\displaystyle d_2\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}State the distance formula
\displaystyle =\displaystyle \sqrt{\left(4-\left(-4\right)\right)^2+\left(10-1\right)^2}Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(8\right)^2+\left(9\right)^2}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{64+81}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{145}Evaluate the addition

12\neq \sqrt{145}, so this point does not lie on the parabola.

Reflect and check
-8
-7
-6
-5
-4
-3
-2
-1
1
2
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x
-2
-1
1
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y

If we look at the graph, we see the point is very close to being on the parabola. Sometimes it may appear that a point lies on the graph, but we should use algebra to verify.

c

\left(-10,5.5\right)

Worked Solution
Create a strategy

We can use the same approach as we did in part (b).

  1. Find the distance of the point to the directrix using absolute values
  2. Find the distance of the point to the focus using the distance formula
Apply the idea

The distance from the point to the directrix is

d_1=\left|5.5-\left(-2\right)\right|=7.5

The distance from the point to the focus is

\displaystyle d_2\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}State the distance formula
\displaystyle =\displaystyle \sqrt{\left(-10-\left(-4\right)\right)^2+\left(5.5-1\right)^2}Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(-6\right)^2+\left(4.5\right)^2}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{36+20.25}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{56.25}Evaluate the addition
\displaystyle =\displaystyle 7.5Evaluate the radical

d_1=d_2 which means this point does lie on the parabola.

Example 2

The point \left(-2, y\right) lies on a parabola. The focus of the parabola is at \left(-4,-1\right) and the directrix is the x-axis. Find the y-coordinate of the point.

Worked Solution
Create a strategy

Since the focus is below the directrix, we know this parabola will open downward. To find the distance from the point to the directrix, we need to find the absolute value of the difference of the y-values. Then, we can use the distance formula and solve for y.

Apply the idea

The directrix is at the x-axis, so the equation of the directrix is y=0. The distance from the point to the directrix is

d_1=\left|y-0\right|=\left|y\right|

We know the distance from the point to the focus will be equal to the distance from the point to the directrix, so we know d_2=\left|y\right|.

\displaystyle d_2\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}State the distance formula
\displaystyle \left|y\right|\displaystyle =\displaystyle \sqrt{\left(-4-\left(-2\right)\right)^2+\left(-1-y\right)^2}Substitute the coordinates
\displaystyle \left|y\right|\displaystyle =\displaystyle \sqrt{\left(-2\right)^2+\left(-y-1\right)^2}Evaluate the subtraction
\displaystyle \left|y\right|\displaystyle =\displaystyle \sqrt{4+y^2+2y+1}Evaluate the exponents
\displaystyle y^2\displaystyle =\displaystyle y^2+2y+5Square both sides
\displaystyle 0\displaystyle =\displaystyle 2y+5Subtract y^2 from both sides
\displaystyle -2y\displaystyle =\displaystyle 5Subtract 2y from both sides
\displaystyle y\displaystyle =\displaystyle -\dfrac{5}{2}Divide both sides by -2

y=-\dfrac{5}{2}=-2.5

Reflect and check
-6
-4
-2
x
-3
-2
-1
y

This is a graph of the parabola, the focus, the directrix, and the point \left(-2,-2.5\right) that we found.

Idea summary

When given a focus and directrix for a parabola, we can determine if a point lies on the parabola by comparing the distance between the point and the focus to the distance between the point and the directrix. The point will lie on the parabola only if the distances are the same.

Equations of parabolas

Exploration

Consider the parabola:

1
2
3
4
5
6
7
8
x
1
2
3
4
5
6
7
8
y
  1. Verify the point \left(8,7\right) lies on the parabola.
  2. Write an equation that would allow you to find any point, \left(x,y\right), on the parabola.
  3. Solve your equation for y.
  4. Compare and contrast the equation from question 2 to the equation from question 3.

When given a focus and directrix, we can find the equation of any parabola by finding the distance from any point (x,y) to the directrix and setting that equal to the distance from the point to the focus. Some algebraic manipulation will give us the standard form of the equation of the parabola.

\displaystyle \left(x-h\right)^2=4p\left(y-k\right)
\bm{\left(x,y\right)}
coordinates of any point on the parabola
\bm{\left(h,k\right)}
coordinates of the vertex
\bm{\left|p\right|}
focal length
Focal length

The distance from the vertex to the focus or from the vertex to the directrix, denoted by \left|p\right|. Also half the distance from the focus to the directrix.

Standard form of the equation of the parabola highlights key features that we can use to find the equation from the graph or to find the equation when we do not know both the focus and directrix. If we solve the equation for y, we get the equation in vertex form.

\displaystyle y=\frac{1}{4p}\left(x-h\right)^2+k
\bm{\left(x,y\right)}
coordinates of any point on the parabola
\bm{\left(h,k\right)}
coordinates of the vertex
\bm{\left|p\right|}
focal length

Notice that both forms of the equation highlight the coordinates of the vertex and the focal length. In standard form, we find the focal length from the constant multiplied to \left(y-k\right). In vertex form, we find the focal length in the denominator of the constant multiplied to \left(x-h\right)^2.

Other key features we can use to help us find the equation are the locations of the focus and directrix and the sign of p. When the focus lies above the directrix, the parabola opens upward and p is positive. When the focus lies below the directrix, the parabola opens downward and p is negative.

x
y
p is positive
x
y
p is negative

Recall from Algebra 1 lesson  10.03 Quadratic functions in vertex form  that we can complete the square with an equation in the form y=ax^2+bx+c to get it into vertex form by following these steps:

  1. Ensure y is on the left side of the equation and the right side is in the form ax^2+bx+c.

  2. Complete the square by adding and subtracting \left(\dfrac{b}{2}\right)^2 to the right side of the equation.

  3. Factor the perfect square trinomial.

  4. Evaluate the addition or subtraction to find k.

Examples

Example 3

Derive the equation of the parabola with a focus at \left(0,-3.5\right) and a directrix at y=1.5. State the equation in vertex form.

Worked Solution
Create a strategy

The equation of a parabola can be found by setting the distance from the point to the directrix equal to the distance from the point to the focus. Since we need to represent all points on the parabola, we will let the point be \left(x,y\right).

Apply the idea

The distance from the point to the directrix is \left|y-1.5\right|

The distance from the point to the focus is \sqrt{\left(x-0\right)^2+\left(y-\left(-3.5\right)\right)^2}=\sqrt{x^2+\left(y+3.5\right)^2}

Setting these distances equal to each other, we get

\displaystyle \left|y-1.5\right|\displaystyle =\displaystyle \sqrt{x^2+\left(y+3.5\right)^2}Definition of parabola
\displaystyle \left(y-1.5\right)^2\displaystyle =\displaystyle x^2+\left(y+3.5\right)^2Square both sides
\displaystyle y^2-3y+2.25\displaystyle =\displaystyle x^2+y^2+7y+12.25Evaluate the exponents
\displaystyle -10y\displaystyle =\displaystyle x^2+10Subtract both sides by y^2, 7y and 2.25
\displaystyle y\displaystyle =\displaystyle -\dfrac{1}{10}x^2-1Divide both sides by -10
Reflect and check

The vertex of this equation is at \left(0, -1\right). It may be easier to see if we rewrite it as y=-\dfrac{1}{10}\left(x-0\right)^2-1

Example 4

Use the graph to answer the following questions. The focus is marked by a point, and the directrix is drawn as a dashed line.

-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
-18
-16
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-12
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a

Identify the vertex.

Worked Solution
Create a strategy

It is important to consider the labels of the graph when looking for coordinates. For this graph, the axes are labeled by twos.

Apply the idea

The vertex is \left(-11, -12\right).

b

Identify \left|p\right|, the focal length.

Worked Solution
Create a strategy

The focal length is the distance from the vertex to the focus. It is also the distance from the vertex to the directrix.

Apply the idea

On the graph, we see that the y-value of the focus is -10, and the y-value of the vertex is -12. This means \left|p\right|=\left|-10-\left(-12\right)\right|=2.

Reflect and check

To verify our answer, we can make sure this is the same distance as the vertex to the directrix. \left|p\right|=\left|-12-\left(-14\right)\right|=2

c

State the equation of the parabola in standard form.

Worked Solution
Create a strategy

When we write the equation of the parabola, we need to pay attention to the sign of p. We know that \left|p\right|=2, so p could be either 2 or -2. When the parabola is facing upward, as it is in this graph, p is positive, so p=2.

Apply the idea

In the previous parts, we found h=-11, k=-12 and p=2.

\displaystyle \left(x-h\right)^2\displaystyle =\displaystyle 4p\left(y-k\right)Standard form of the equation
\displaystyle \left(x-\left(-11\right)\right)^2\displaystyle =\displaystyle 4\cdot 2\left(y-\left(-12\right)\right)Substitute the coordinates
\displaystyle \left(x+11\right)^2\displaystyle =\displaystyle 8\left(y+12\right)Evaluate the multiplication
Reflect and check

We can solve the equation for y to put this into vertex form.

\displaystyle \left(x+11\right)^2\displaystyle =\displaystyle 8\left(y+12\right)Standard form
\displaystyle \dfrac{1}{8}\left(x+11\right)^2\displaystyle =\displaystyle y+12Divide both sides by 8
\displaystyle \dfrac{1}{8}\left(x+11\right)^2-12\displaystyle =\displaystyle ySubtract 12 from both sides

The equation in vertex form is y=\dfrac{1}{8}\left(x+11\right)^2-12.

Example 5

The equation of a parabola is 0=x^2+3x+\dfrac{1}{2}y-4.

a

Rewrite the equation in vertex form.

Worked Solution
Create a strategy

To write the equation in vertex form, we need to complete the square. To complete the square we:

  1. Move the y-term to the left side so the right side of the equation has an expression in the form ax^{2} + bx +c
  2. Complete the square by adding and subtracting \left(\frac{b}{2}\right)^2 to the right side of the equation
  3. Factor the perfect square trinomial

After this, we will need to solve for y in order to get the entire equation in vertex form.

Apply the idea

First, we will move the y-term to the other side. -\dfrac{1}{2}y=x^2+3x-4

To complete the square, we need to add and subtract \left(\dfrac{b}{2}\right)^2 to the right side of the equation. In this quadratic, b=3: \left(\dfrac{b}{2}\right)^2=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4} This is what we need to add and subtract from the right side in order to complete the square.

\displaystyle -\dfrac{1}{2}y\displaystyle =\displaystyle x^2+3x+\dfrac{9}{4}-4-\dfrac{9}{4}Complete the square
\displaystyle -\dfrac{1}{2}y\displaystyle =\displaystyle \left(x+\dfrac{3}{2}\right)^2-4-\dfrac{9}{4}Factor the perfect square trinomial
\displaystyle -\dfrac{1}{2}y\displaystyle =\displaystyle \left(x+\dfrac{3}{2}\right)^2-\dfrac{25}{4}Evaluate the subtraction
\displaystyle y\displaystyle =\displaystyle -2\left(x+\dfrac{3}{2}\right)^2+\dfrac{25}{2}Multiply both sides by -2
Reflect and check

In this form, we can easily identify the vertex at \left(-\dfrac{3}{2},\dfrac{25}{2}\right).

b

Find \left|p\right|, the focal length.

Worked Solution
Create a strategy

In vertex form, the focal length can be found in the coefficient of \left(x-h\right)^2.

Apply the idea

In the general formula, the coefficient is \dfrac{1}{4p}. In the equation we found, the coefficient is -2. We can set these equal to each other and solve for the value of p.

\displaystyle \dfrac{1}{4p}\displaystyle =\displaystyle -2
\displaystyle 1\displaystyle =\displaystyle -8p
\displaystyle -\dfrac{1}{8}\displaystyle =\displaystyle p

Since the focal length represents distance, we need to take the absolute value of this answer. So, the focal length is \left|-\dfrac{1}{8}\right|=\dfrac{1}{8}. The focus is \dfrac{1}{8} units from the vertex.

Reflect and check

To check our answer, we can substitute p=-\dfrac{1}{8} back into \dfrac{1}{4p}=-2.

\displaystyle \dfrac{1}{4p}\displaystyle =\displaystyle -2
\displaystyle \dfrac{1}{4\cdot -\frac{1}{8}}\displaystyle =\displaystyle -2
\displaystyle \dfrac{1}{-\frac{1}{2}}\displaystyle =\displaystyle -2
\displaystyle -2\displaystyle =\displaystyle -2
c

State the coordinates of the focus.

Worked Solution
Create a strategy

In part (a), we found the vertex was \left(-\dfrac{3}{2},\dfrac{25}{2}\right).In part (b), we said the focus was \dfrac{1}{8} units away. Now, we need to determine if the focus is above or below the vertex.

Apply the idea

The focus of a parabola always lies insides the parabola. We found p to be a negative value in part (b) which tells us the parabola is facing downward. This also tells us the focus is below the vertex.

The focus is \dfrac{1}{8} units below the vertex, so we need to subtract \dfrac{1}{8} from the y-value of the vertex.\dfrac{25}{2}-\dfrac{1}{8}=\dfrac{99}{8}

The x-value of the focus is the same as the x-value of the vertex, so the coordinates of the focus are \left(-\dfrac{3}{2},\dfrac{99}{8}\right).

Reflect and check

Both the focus and the vertex lie on the line of symmetry which is x=h. For this parabola, the line of symmetry is x=-\dfrac{3}{2}.

d

State the equation of the directrix.

Worked Solution
Create a strategy

The directrix will be on the opposite side of the vertex as the focus. Since the focus was \dfrac{1}{8} units below the vertex, the directrix will be \dfrac{1}{8} units above the vertex.

Apply the idea

This time, we need to add \dfrac{1}{8} to the y-value of the vertex. \dfrac{25}{2}+\dfrac{1}{8}=\dfrac{101}{8}

The directrix is a horizontal line, so we need to state our answer as an equation in the form y=c. The directrix is y=\dfrac{101}{8}.

Reflect and check

This parabola and its key features are graphed below. We need to graph this within a small range of values in order to see the focal length because it is such a small value.

-2
-1
x
11
12
13
y
Idea summary

We can determine the equation of any parabola when given information about its key features. The standard form of the equation of a parabola is

\displaystyle \left(x-h\right)^2=4p\left(y-k\right)
\bm{\left(x,y\right)}
coordinates of any point on the parabola
\bm{\left(h,k\right)}
coordinates of the vertex
\bm{\left|p\right|}
focal length

When the parabola opens upward, p is positive. When the parabola opens downward, p is negative.

Outcomes

G.GPE.A.2

Derive the equation of a parabola given a focus and directrix.

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