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6.7: Hyperbolas and Asymptotes

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Like other conic sections, hyperbolas can be created by "slicing" a cone and looking at the cross-section. Unlike other conics, hyperbolas actually require 2 cones stacked on top of each other, point to point. The shape is the result of effectively creating a parabola out of both cones at the same time.

So the question is, should hyperbolas really be considered a shape all their own? Or are they just two parabolas graphed at the same time? Could "different" shapes be made from any of the other conic sections if two cones were used at the same time?

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Guidance

In addition to their focal property, hyperbolas also have another interesting geometric property. Unlike a parabola, a hyperbola becomes infinitesimally close to a certain line as the $x-$ or $y-$ coordinates approach infinity. What we mean by “infinitesimally close?” Here we mean two things: 1) The further you go along the curve, the closer you get to the asymptote, and 2) If you name a distance, no matter how small, eventually the curve will be that close to the asymptote. Or, using the language of limits, as we go further from the vertex of the hyperbola the limit of the distance between the hyperbola and the asymptote is 0.

These lines are called asymptotes .

There are two asymptotes, and they cross at the point at which the hyperbola is centered:

For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , the asymptotes are the lines:

$y = \frac{b}{a} x$ and $y = - \frac{b}{a} x$ .

For a hyperbola of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the asymptotes are the lines:

$y = \frac{a}{b} x$ and $y = - \frac{a}{b} x$ .

(For a shifted hyperbola, the asymptotes shift accordingly.)

Example A

Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better drawing: $9x^2 - 36x - 4y^2 - 16y - 16 = 0$ .

Solution

First, we put the hyperbola into the standard form:

$9(x^2 - 4x) - 4(y^2 + 4y) & = 16 \\9(x^2 - 4x + 4) - 4(y^2 + 4y + 4) & = 36 \\\frac{(x - 2)^2}{4} - \frac{(y + 2)^2}{9} & = 1$

So $a = 2$ , $b =3$ and $c = \sqrt{4 + 9}=\sqrt{13}$ . The hyperbola is horizontally oriented, centered at the point (2,-2), with foci at $\left (2 + \sqrt{13}, -2 \right )$ and $\left (2 - \sqrt{13}, -2 \right )$ . After taking shifting into consideration, the asymptotes are the lines: $y + 2 = \frac{3}{2} (x - 2)$ and $y + 2 = - \frac{3}{2} (x - 2)$ . So graphing the vertices and a few points on either side, we see the hyperbola looks something like this:

Example B

Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better drawing: $16x^2 - 96x - 9y^2 - 36y - 84 = 0$

Solution

Example C

Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better drawing: $y^2 - 14y -25x^2 - 200x -376 = 0$

Solution

Concept question wrap-up

Hyperbolas are considered different shapes, because there are specific behaviors that are unique to hyperbolas. Also, though hyperbolas are the result of dual parabolas, none of the other conics really create unique shapes with dual cones - just double figures - and in any case require multiple "slices".

Vocabulary

A hyperbola is a conic section where the cutting plane intersects both sides of the cone, resulting in two infinite “U”-shapes curves.

An unbounded shape is so large that no circle, no matter how large, can enclose the shape.

An asymptote is a line which a curve approaches as the curve and the line approach infinity, eventually becoming closer than any given positive number.

A perpendicular hyperbola is a hyperbola where the asymptotes are perpendicular.

Guided Practice

Questions

1) Find the equation for a hyperbola with asymptotes of slopes $\frac{5}{12}$ and $- \frac{5}{12}$ , and foci at points (2,11) and (2,1).

2) A hyperbola with perpendicular asymptotes is called perpendicular . What does the equation of a perpendicular hyperbola look like?

3) Find an equation of the hyperbola with x-intercepts at x = –7 and x = 5, and foci at (–6, 0) and (4, 0).

Solutions

1) $\frac{(y - 6)^2}{25} - \frac{(x - 2)^2}{144} = 1$

2) The slopes of perpendicular lines are negative reciprocals of each other. This means that $\frac{a}{b} = \frac{b}{a}$ , which, for positive $a$ and $b$ means $a = b$

3) To find the equation:

The foci have the same y-coordinates, so this is a left/right hyperbola with the center, foci, and vertices on a line paralleling the x-axis.
Since it is a left/right hyperbola, the y part of the equation will be negative and equation will lead with the $x^2$ term (since the leading term is positive by convention and the squared term must have different signs if this is a hyperbola). :The center is midway between the foci, so the center $(h, k) = (-1, 0)$ . The foci c are 5 units to either side of the center, so $c = 5 \to c^2 = 25$
The x-intercepts are 4 units to either side of the center, and the foci are on the x-axis so the intercepts must be the vertices a $a = 4 \to a^2 = 16$
Use Pythagoras $a^2 + b^2 = c^2$ to get $b^2 = 25 - 16 = 9$

Substitute the calculated values into the standard form $\frac{(x - h)^2}{a} - \frac{(y - k)^2}{b} = 1$ to get

$\frac{(x + 1)^2}{16} - \frac{y^2}{9} = 1$

Practice

Find the equations of the asymptotes

1. $\frac{(y + 3)^2}{4} - (x - 2)^2 = 1$
2. $\frac{y^2}{16} - (x + 3)^2 = 1$
3. $\frac{(x + 2)^2}{4} - \frac{(y + 1)^2}{9} = 1$
4. $\frac{(y - 4)^2}{16} - \frac{(x - 4)^2}{16} = 1$
5. $\frac{(x - 1)^2}{1} - \frac{9(y + 4)^2}{1} = 9$
6. $\frac{(y + 2)^2}{16} - \frac{(x - 2)^2}{1} = 1$
7. $\frac{(x - 4)^2}{1} - \frac{(y + 1)^2}{4} = 1$
8. $\frac{y^2}{16} - \frac{(x + 1)^2}{4} = 1$
9. $\frac{(x - 3)^2}{4} - \frac{(y - 4)^2}{1} = 1$
10. $\frac{(x - 4)^2}{4} - \frac{(y - 3)^2}{1} = 1$

Graph the hyperbolas, give the equation of the asymptotes, use the asymptotes to enhance the accuracy of your graph.

1. $\frac{(x + 4)^2}{4} - \frac{(y - 1)^2}{9} = 1$
2. $\frac{(y + 3)^2}{4} - \frac{(x - 4)^2}{9} = 1$
3. $\frac{(y + 4)^2}{16} - \frac{(x - 1)^2}{4} = 1$
4. $(x - 2)^2 - 4y^2 = 16$
5. $\frac{y^2}{4} - \frac{(x - 1)^2}{4} = 1$
6. $\frac{(x - 2)^2}{16} - \frac{(y + 4)^2}{1} = 1$
7. $\frac{(x + 2)^2}{9} - \frac{(y + 2)^2}{16} = 1$
8. $\frac{(x + 4)^2}{9} - \frac{(y - 2)^2}{4} = 1$

Nov 01, 2012

Aug 13, 2014

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