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High School Algebra II: Tutoring Solution26 chapters | 274 lessons

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Lesson Transcript

Instructor:
*David Karsner*

Hyperbolas are a set of two similar curves called branches set on a plane governed by certain equations. They are, along with circles, parabolas and ellipses, conic sections, as you'll learn in this lesson.

If you were to stand on the ground and throw a ball into the air, the ball would leave your hand reach a certain height and then fall back to the ground. The path that the ball would take would look like an arc. If this path were visible and you could somehow hold it up to a giant mirror, then the path and its reflection would create hyperbola. This lesson will give you the method in which one can take an equation of a hyperbola and find its center, vertices, and asymptotes and then graph it.

In this image we can see how a **hyperbola** is created from the intersection of a plane and two cones that meet on their tips. A hyperbola can open to the left and right or open up and down. A more formal definition of a **hyperbola** is a collection of all points, whose distances to two fixed points, called **foci** (plural for focus), is a constant difference. The equation of a hyperbola in standard form is:

((*x* - *h*)^2 / *a*^2) - ((*y* - *k*)^2 / *b*^2) = 1

The *x* and *y* are interchangeable and both give you an equation of an hyperbola. Let's look at some of its parts.

Notice how the hyperbola has two lines of symmetry: one vertical and one horizontal. You can fold the hyperbola so that one half of it will lie completely on top of the other half. The folds will create two lines. The point at which the vertical fold and horizontal fold intersect is called the **center** of the hyperbola. Using the standard form of the hyperbola, the center is located at the point (*h*, *k*).

The vertex of a parabola is the lowest point on a parabola if it is opening up and the highest point if it is opening down. The **vertices** of a hyperbola (which is composed of two parabolas) is the vertex of each branch of the hyperbola. To use the standard form of the hyperbola to find the vertices, you need to notice if the positive term is *x*^2 or *y*^2. If the positive term is *x*^2, the vertices are found at (*h* + *a*, *k*) and (*h* - *a*, *k*). If *y*^2 is the first term, the vertices are found at (*h*, *k* + *b*) and (*h*, *k* - *b*).

The **asymptotes** of a hyperbola are two lines that lie between the two branches of the hyperbola. The intersection of the two lines occurs at the center of the hyperbola. The lines come very close to the hyperbola itself but don't cross it. If the equation has a positive *x*^2 term, then equations for the lines will be:

- (
*y*-*k*) = +*b*/*a*(*x*-*h*) - (
*y*-*k*) = -*b*/*a*(*x*-*h*)

If the *y*^2 term is positive, the equations will be the same, except *b*/*a* is replaced with *a*/*b*.

In the standard form of the equation, the values for *a*, *b*, *g*, and *k* are accessible. Knowing these values will make graphing the hyperbola possible. Converting the equation of a hyperbola so that it is in the standard form requires the ability to **complete the square**.

Put the following equation into standard form:

4*x*^2 - *y*^2 - 24*x* + 4*y* + 28 = 0

Group the terms so that the *x* terms are together, the *y* terms are together, and the constant is on the right side of the equation:

(4*x*^2 - 24*x*) + (-*y*^2 + 4*y*) = -28

Factor out any common factors from each grouping:

4(*x*^2 - 6*x*) - 1(*y*^2 - 4*y*) = -28

Add whatever is needed to each grouping to complete the square:

4(*x*^2 - 6*x* + 9) - 1(*y*^2 - 4*y* + 4) = -28 + 4(9) - 1(4)

Factor and simplify:

4(*x* - 3)^2 - 1(*y* - 2)^2 = 4

Divide the equation by the factors in front of the groupings:

((*x* - 3)^2) / 1 - ((*y* - 2)^2) / 4 = 1

Using the standard form:

*h*is 3*k*is 2*a*is 1*b*is 2

Step One: To graph a hyperbola, you will need to determine the values of *h*, *k*, *a*, and *b* by putting the equation into the standard form, if it's not already.

Step Two: Find and draw the center (*h*, *k*)

Step Three: Draw the vertices of the hyperbola

Step Four: Draw the asymptotes. These two lines should intersect each other at the center.

Step Five: Draw the two branches of the hyperbola that come close to the asymptotes and cross through the vertex.

25*x*^2 - 16*y*^2 - 100*x* - 32*y* - 316 = 0

((*x* - 2)^2 / 16) - ((*y* + 1)^2 / 25 = 1

*h* = 2, *k* = -1, *a* = 4, *b* = 5

(*h*, *k*) is (2, -1)

The *x*^2 is positive so the vertices are found at (*h* + *a*, *k*) and (*h* - *a*, *k*)

(6, -1)( -2, -1)

The *x*^2 is positive, so the slope is *b*/*a*, -*b*/*a*, which gives us 5/4 and -5/4, and it goes through the center (2, -1)

**Hyberbolas** are one of the conic sections. That means that they are created when you slice a plane through two cones. The standard form of a hyperbola is:

(( *x* - *h*)^2/*a*^2) - ((*y* - *k*)^2/*b*^2)

In this form you can determine the *h*, *k*, *a*, and *b* which allows one to find the center, vertices, and asymptotes of a hyperbola. It also allows you to graph a hyperbola.

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High School Algebra II: Tutoring Solution26 chapters | 274 lessons

- Defining and Graphing Ellipses in Algebra 5:00
- How to Write the Equation of an Ellipse in Standard Form 6:18
- The Circle: Definition, Conic Sections & Distance Formula 3:43
- The Hyperbola: Definition, Vertices, Foci & Graphing 10:00
- How to Write the Equation of a Hyperbola in Standard Form 8:14
- The Parabola: Definition & Graphing 8:33
- How to Write the Equation of a Parabola in Standard Form 8:17
- How to Identify a Conic Section by Its Formulas 6:33
- Hyperbola: Definition, Formula & Examples 6:23
- Go to Algebra II - Conic Sections: Tutoring Solution

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