Lesson Objectives

- Learn how to find the intercepts for a hyperbola
- Learn how to sketch the fundamental rectangle for a hyperbola
- Learn how to sketch the asymptotes for a hyperbola
- Learn how to sketch the graph of a hyperbola

## How to Graph a Hyperbola

In the last lesson, we learned how to graph an ellipse. In this lesson, we will learn how to graph a hyperbola that is centered at the origin. A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points is constant. The equation for the hyperbola looks similar to that of an ellipse. The difference is the minus sign in the middle.

When we work with a hyperbola that is centered at the origin, we will see two different equations.

(a, 0) and (-a, 0)

This type of hyperbola opens left and right. We will not have any y-intercepts. To think about why let's plug a 0 in for x and solve for y: $$\frac{(0)^2}{a^2}- \frac{y^2}{b^2}=1$$ $$- \frac{y^2}{b^2}=1$$ $$-y^2=b^2$$ Let's multiply both sides by -1: $$y^{2}=-1 \cdot b^{2}$$ To solve for y, let's take the square root of each side: $$y=\pm \sqrt{-1 \cdot b^2}$$ $$y=\pm bi$$ Since this problem will not have a real solution, we will not have any y-intercepts.

Our y-intercepts will occur at:

(0, b) and (0, -b)

This type of hyperbola opens up and down. We will not have any x-intercepts. To think about why let's plug a 0 in for y and solve for x: $$\frac{(0)^2}{b^2}- \frac{x^2}{a^2}=1$$ $$- \frac{x^2}{a^2}=1$$ $$-x^2=a^2$$ Let's multiply both sides by -1: $$x^{2}=-1 \cdot a^{2}$$ To solve for x, let's take the square root of each side: $$x=\pm \sqrt{-1 \cdot a^2}$$ $$x=\pm ai$$ Again, we know this problem will not have a real solution, therefore, we will not have any x-intercepts.

To graph a hyperbola:

Example 1: Sketch the graph of each hyperbola. $$\frac{y^2}{4}- \frac{x^2}{9}=1$$ After inspecting the equation, we can determine this is the graph of a hyperbola that is centered at the origin (0, 0) and opens up and down. This means we will have y-intercepts and no x-intercepts.

To graph our hyperbola, we can follow the above steps. We will begin by plotting the intercepts. To find these, take the principal and negative square root of the number underneath y

(3,2),(-3,2),(-3,-2),(3,-2). Now we can sketch the asymptotes. These have the following equations: $$y=\frac{2}{3}x$$ $$y=-\frac{2}{3}x$$ Lastly, we can sketch the graph of the hyperbola. One branch opens up and the other opens down. Each branch will go through the y-intercept and approach but not touch the asymptotes. Let's look at another example.

Example 2: Sketch the graph of each hyperbola. $$\frac{x^{2}}{4}- \frac{y^{2}}{4}=1$$ After inspecting the equation, we can determine this is the graph of a hyperbola that is centered at the origin (0, 0) and opens left and right. This means we will have x-intercepts and no y-intercepts. To graph our hyperbola, we can follow the above steps. We will begin by plotting the intercepts. To find these, take the principal and negative square root of the number underneath x

When we work with a hyperbola that is centered at the origin, we will see two different equations.

### A Hyperbola with x-intercepts (a,0) and (-a,0)

$$\frac{x^2}{a^2}- \frac{y^2}{b^2}=1$$ Our x-intercepts will occur at:(a, 0) and (-a, 0)

This type of hyperbola opens left and right. We will not have any y-intercepts. To think about why let's plug a 0 in for x and solve for y: $$\frac{(0)^2}{a^2}- \frac{y^2}{b^2}=1$$ $$- \frac{y^2}{b^2}=1$$ $$-y^2=b^2$$ Let's multiply both sides by -1: $$y^{2}=-1 \cdot b^{2}$$ To solve for y, let's take the square root of each side: $$y=\pm \sqrt{-1 \cdot b^2}$$ $$y=\pm bi$$ Since this problem will not have a real solution, we will not have any y-intercepts.

### A Hyperbola with y-intercepts (0,b) and (0,-b)

$$\frac{y^2}{b^2}- \frac{x^2}{a^2}=1$$ Note: Some textbooks will keep a^{2}under the y^{2}. Keeping b^{2}underneath y^{2}will make some formulas for this lesson a bit easier to remember.Our y-intercepts will occur at:

(0, b) and (0, -b)

This type of hyperbola opens up and down. We will not have any x-intercepts. To think about why let's plug a 0 in for y and solve for x: $$\frac{(0)^2}{b^2}- \frac{x^2}{a^2}=1$$ $$- \frac{x^2}{a^2}=1$$ $$-x^2=a^2$$ Let's multiply both sides by -1: $$x^{2}=-1 \cdot a^{2}$$ To solve for x, let's take the square root of each side: $$x=\pm \sqrt{-1 \cdot a^2}$$ $$x=\pm ai$$ Again, we know this problem will not have a real solution, therefore, we will not have any x-intercepts.

To graph a hyperbola:

- Locate and plot the intercepts
- (a, 0) and (-a, 0) if the x
^{2}term has a positive coefficient - (0, b) and (0, -b) if the y
^{2}term has a positive coefficient

- (a, 0) and (-a, 0) if the x
- Find the fundamental rectangle
- Endpoints: (a,b),(-a,b),(-a,-b),(a,-b)

- Sketch the asymptotes
- These are extended from the fundamental rectangle
- y = (b/a)x
- y = -(b/a)x

- Sketch the graph: each branch goes through an intercept and approaches but doesn't touch the asymptote

Example 1: Sketch the graph of each hyperbola. $$\frac{y^2}{4}- \frac{x^2}{9}=1$$ After inspecting the equation, we can determine this is the graph of a hyperbola that is centered at the origin (0, 0) and opens up and down. This means we will have y-intercepts and no x-intercepts.

To graph our hyperbola, we can follow the above steps. We will begin by plotting the intercepts. To find these, take the principal and negative square root of the number underneath y

^{2}. This will give us 2 and -2, therefore, our y-intercepts will occur at: (0, 2) and (0, -2). Now we will find the fundamental rectangle. The endpoints occur at:(3,2),(-3,2),(-3,-2),(3,-2). Now we can sketch the asymptotes. These have the following equations: $$y=\frac{2}{3}x$$ $$y=-\frac{2}{3}x$$ Lastly, we can sketch the graph of the hyperbola. One branch opens up and the other opens down. Each branch will go through the y-intercept and approach but not touch the asymptotes. Let's look at another example.

Example 2: Sketch the graph of each hyperbola. $$\frac{x^{2}}{4}- \frac{y^{2}}{4}=1$$ After inspecting the equation, we can determine this is the graph of a hyperbola that is centered at the origin (0, 0) and opens left and right. This means we will have x-intercepts and no y-intercepts. To graph our hyperbola, we can follow the above steps. We will begin by plotting the intercepts. To find these, take the principal and negative square root of the number underneath x

^{2}. This will give us 2 and -2, therefore, our x-intercepts will occur at: (2, 0) and (-2, 0). Now we will find the fundamental rectangle. The endpoints occur at: (2,2),(-2,2),(-2,-2),(2,-2). Now we can sketch the asymptotes. These have the following equations: $$y=x$$ $$y=-x$$ Lastly, we can sketch the graph of the hyperbola. One branch opens left and the other opens right. Each branch will go through the x-intercept and approach but not touch the asymptotes.#### Skills Check:

Example #1

Find the endpoints of the fundamental rectangle. $$\frac{x^{2}}{16}- y^{2}=1$$

Please choose the best answer.

A

$$(1, 4), (-1, 4), (1, -4), (-1, -4)$$

B

$$(2, 4), (-2, 4), (2, -4), (-2, -4)$$

C

$$(4, 1), (-4, 1), (4, -1), (-4, -1)$$

D

$$(4, 3), (-4, 3), (4, -3), (-4, -3)$$

E

$$(6, 3), (-6, 3), (6, -3), (-6, -3)$$

Example #2

Find the x-intercepts. $$\frac{y^{2}}{36}- \frac{x^{2}}{81}=1$$

Please choose the best answer.

A

$$(6, 0), (-6, 0)$$

B

$$\text{None}$$

C

$$(\sqrt{6}, 0), (-\sqrt{6}, 0)$$

D

$$(9, 0), (-9, 0)$$

E

$$(3, 0), (-3, 0)$$

Example #3

Find the asymptotes. $$\frac{y^{2}}{100}- \frac{x^{2}}{4}=1$$

Please choose the best answer.

A

$$y=\pm 10x$$

B

$$y=\pm \sqrt{10}x$$

C

$$y=\pm \frac{2}{3}x$$

D

$$y=\pm \frac{25}{2}x$$

E

$$y=\pm 5x$$

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test