Precalculus by Richard Wright

And surely I am with you always, to the very end of the age. Matthew 28:20b NIV

Summary: In this section, you will:

- Find the foci, vertices, covertices, and asymptotes of a hyperbola.
- Write the standard equation of a hyperbola.
- Graph a hyperbola.
- Classify conics based on the general equation.

SDA NAD Content Standards (2018): PC.6.7

When designing cooling towers for power plants, engineers needed to find a shape that withstands high winds and is built with as little material as possible. A hyperbolic shape solves both problems. In addition, the shape helps the cooling air naturally rise.

Hyperbolas are the set of all points in a plane where the difference of the distances from two set points, foci, is constant. In figure 2, *d*_{1} – *d*_{2} is constant.

The vertices are the minimum/maximum of each branch of the hyperbola. The covertices are analogous to the covertices of the ellipse, but are not actually on the hyperbola. The transverse axis is the segment connecting the vertices, and the conjugate axis is the segment connecting the covertices. A box with the vertices and covertices as midpoints of the sides provides an aid for sketching the graph of the hyperbola. The branches are asymptotic with the asymptotes going through the corners of the box.

## Horizontal Hyperbola |
## Vertical Hyperbola |

Center: (h, k)Horizontal transverse axis length = 2 aVertical conjugate axis length = 2 bc^{2} = a^{2} + b^{2}Vertices: ( h ± a, k)Covertices: ( h, k ± b)Foci: ( h ± c, k)Asymptotes: \(y = k ± \frac{b}{a}(x - h)\) Standard Equation: \(\frac{\left(x - h\right)^2}{a^2} - \frac{\left(y - k\right)^2}{b^2} = 1\) |
Center at (h, k)Vertical transverse axis length = 2 aHorizontal conjugate axis length = 2 bc^{2} = a^{2} + b^{2}Vertices ( h, k ± a)Covertices ( h ± b, k)Foci ( h, k ± c)Asymptotes: \(y = k ± \frac{a}{b}(x - h)\) Standard Equation: \(\frac{\left(y - k\right)^2}{a^2} - \frac{\left(x - h\right)^2}{b^2} = 1\) |

Eccentricity is \(e = \frac{c}{a}\) where *e* > 1. The larger the eccentricity is, the closer the branches are to straight lines.

Find the center, vertices, asymptotes, and foci of the hyperbola given by 16*x*^{2} − 4*y*^{2} = 64.

Write the equation in standard form by dividing by 64 so that the equation equals 1.

$$\frac{x^2}{4} - \frac{y^2}{16} = 1$$

Because *x* comes first, this is a horizontal hyperbola. Compare this equation with the standard form, \(\frac{\left(x - h\right)^2}{a^2} - \frac{\left(y - k\right)^2}{b^2} = 1\), to see that *h* = 0, *k* = 0, *a*^{2} = 9 so *a* = 3, and *b*^{2} = 4 so *b* = 2.

*c*^{2} = *a*^{2} + *b*^{2}

*c*^{2} = 4 + 16 = 20

$$c = 2\sqrt{5}$$

Center is (0, 0).

Vertices are (±2, 0).

Covertices are (0, ±4).

Asymptotes are *y* = ±2*x*.

Foci are (\(±2\sqrt{5}\), 0).

Find the center, vertices, covertices, asymptotes, and foci of \(\frac{(y - 2)^2}{36} - \frac{(x + 5)^2}{49} = 1\).

C(−5, 2), V(−5, −4) and (−5, 8), CV(−12, 2) and (2, 2), Asymptotes: \(y = 2 ± \frac{6}{7}(x + 5)\), F(-5, 2 ± \(\sqrt{85}\))

Find the standard form of the equation of the hyperbola centered at (−3, 1) with transverse axis length 6 and foci at (−3, −4) and (−3, 6). Then graph the hyperbola.

Graphing the three points shows that the transverse axis is vertical.

The transverse axis length is 2*a* = 6, so *a* = 3.

The distance from the center to the foci is *c*, so *c* = 5.

Find *b*

*c*^{2} = *a*^{2} + *b*^{2}

5^{2} = 3^{2} + *b*^{2}

*b*^{2} = 16

*b* = 4

The center is (−3, 1) so h = −3 and k = 1.

Write the standard form equation for a vertical hyperbola.

$$\frac{(y - 1)^2}{9} - \frac{(x + 3)^2}{16} = 1$$

Graph it by plotting the center.

Given that it is a vertical ellipse, move the distance *a* = 5 vertically from the center and a distance *b* = 4 horizontally from the center.

Draw a rectangle with these four points as the center of each side.

Draw diagonal lines through the corners of the rectangle. These are the asymptotes.

Sketch the hyperbola starting near an asymptote, curving through the vertex, and ending near the other asymptote.

Find the standard form of the equation of the hyperbola with vertices at (1, 5) and (1, -3) and foci at (1, 7) and (1, -5). Then graph the hyperbola.

\(\frac{(y - 1)^2}{16} - \frac{(x - 1)^2}{20} = 1\),

Sketch the graph of the hyperbola given by 9*x*^{2} − 25*y*^{2} + 18*x* + 50*y* − 191 = 0.

Write the equation in standard form by completing the square. Start by moving the constant to the other side and factoring the *x*’s and *y*’s.

9(*x*^{2} + 2*x*) − 25(*y*^{2} − 2y) = 191

Add \(\left(\frac{1}{2} b\right)^2\) for both the *x*’s and *y*’s. On the right side, don’t forget to multiply by the coefficient.

$$9\left(x^2 + 2x + \left(\frac{1}{2} \left(2\right)\right)^2\right) - 25\left(y^2 - 2y + \left(\frac{1}{2}\left(-2\right)\right)^2\right) = 191 + 9\left(\frac{1}{2}\left(2\right)\right)^2 - 25\left(\frac{1}{2}\left(-2\right)\right)^2$$

9(*x*^{2} + 2*x* + 1) − 25(*y*^{2} − 2y + 1) = 191 + 9 − 25

Factor the left side.

9(*x* + 1)^{2} − 25(*y* − 1)^{2} = 175

Divide both sides by 175.

$$\frac{(x + 1)^2}{25} - \frac{(y - 1)^2}{9} = 1$$

This is a horizontal hyperbola.

The center is at (−1, 1), *a* = 5 and *b* = 3.

Go horizontally *a* = 5 from the center and vertically *b* = 3 from the center. These four points form the box. Draw the diagonals and then the hyperbola.

Sketch the graph of the hyperbola given by *x*^{2} − 4*y*^{2} + 4*x* = 0.

All vertical and horizontal conics can be written in this general form.

*Ax*^{2} + *Cy*^{2} + *Dx* + *Ey* + *F* = 0

The conics can be classified by the following.

- Circle if
*A*=*C* - Parabola if
*AC*= 0 (so*A*= 0 or*C*= 0) - Ellipse if
*AC*> 0 - Hyperbola if
*AC*< 0

If a conic is written in general form: *Ax*^{2} + *Cy*^{2} + *Dx* + *Ey* + *F* = 0

*A*=*C*→**Circle***AC*= 0 (so*A*= 0 or*C*= 0) →**Parabola***AC*> 0 →**Ellipse***AC*< 0 →**Hyperbola**

Classify each conic.

- 16
*x*^{2}− 3*y*^{2}+ 9*x*−*y*+ 16 = 0 - 7
*y*^{2}+ 2*x*− y + 4 = 0 - 4
*x*^{2}+ 3*y*^{2}− 7*x*+ 2*y*− 1 = 0 - 3
*x*^{2}+ 3*y*^{2}+ 2*x*+*y*+ 4 = 0

Compare each equation to the general equation of a conic, *Ax*^{2} + *Cy*^{2} + *Dx* + *Ey* + *F* = 0, and find *A* and *C*. *A* is the coefficient of the *x*^{2} term and *C* is the coefficient of the *y*^{2} term.

*AC*= 16(−3) = −48 < 0 so it is a hyperbola*AC*= 7(0) = 0 so it is a parabola*AC*= 4(3) = 12 > 0 so it is an ellipse*A*=*C*= 3 so it is a hyperbola

Classify each conic.

- 2
*x*^{2}+ 3*y*^{2}−*x*= 0 - 2
*x*^{2}+ 2*x*− y + 4 = 0

(a) Ellipse; (b) Parabola

## Horizontal Hyperbola |
## Vertical Hyperbola |

Center: (h, k)Horizontal transverse axis length = 2 aVertical conjugate axis length = 2 bc^{2} = a^{2} + b^{2}Vertices: ( h ± a, k)Covertices: ( h, k ± b)Foci: ( h ± c, k)Asymptotes: \(y = k ± \frac{b}{a}(x - h)\) Standard Equation: \(\frac{\left(x - h\right)^2}{a^2} - \frac{\left(y - k\right)^2}{b^2} = 1\) |
Center at (h, k)Vertical transverse axis length = 2 aHorizontal conjugate axis length = 2 bc^{2} = a^{2} + b^{2}Vertices ( h, k ± a)Covertices ( h ± b, k)Foci ( h, k ± c)Asymptotes: \(y = k ± \frac{a}{b}(x - h)\) Standard Equation: \(\frac{\left(y - k\right)^2}{a^2} - \frac{\left(x - h\right)^2}{b^2} = 1\) |

If a conic is written in general form: *Ax*^{2} + *Cy*^{2} + *Dx* + *Ey* + *F* = 0

*A*=*C*→**Circle***AC*= 0 (so*A*= 0 or*C*= 0) →**Parabola***AC*> 0 →**Ellipse***AC*< 0 →**Hyperbola**

Helpful videos about this lesson.

- What is the difference between the transverse axis and the conjugate axis? How are they related to the major and minor axes of an ellipse?
- \(\frac{y^2}{64} - \frac{x^2}{16} = 1\)
- 20
*x*^{2}− 25*y*^{2}− 200 = 0 - \(\frac{(x + 6)^2}{9} - \frac{(y - 3)^2}{8} = 1\)
- 100
*x*^{2}− 81*y*^{2}+ 200*x*+ 648*y*+ 6904 = 0 - Foci: (±4, 0), Vertices: (±2, 0)
- Vertices: (3, 4) and (3, 10), Covertices: (−1, 7) and (7, 7)
- Asymptotes: \(y = 3 ± \frac{2}{3} (x - 1)\), Vertex: (1, 7)
- \(\frac{x^2}{9} - \frac{y^2}{16} = 1\)
*x*^{2}−*y*^{2}− 6*x*+ 4*y*+ 9 = 0- 4
*x*^{2}−*y*^{2}+ 8*x*+ 8*y*− 16 = 0 - Find the standard equation of the hyperbola with \(e = \frac{4}{3}\) and vertices (±3, 0).
- What is the eccentricity of 16
*x*^{2}− 9*y*^{2}+ 96*x*+ 36*y*− 36 = 0? - The cooling tower at the electrical power generating station in Michigan City, Indiana, is modeled by a hyperbola and is about 300 feet tall. Write a model for the sides of the tower if the center is at (0, 200), the vertices are (±75, 200), and point at the base is (112.5, 0).
- A sundial is made of a rod that casts a shadow. The shadow falls on a scale to tell the time. The tip of the rod traces a hyperbola over the course of a day. This is called the declination line. If a certain declination line is modeled by \(\frac{y^2}{4} - \frac{x^2}{100} = 1\), what is its eccentricity?
- (7-03) Find the center, vertices, covertices, and foci of \(\frac{(x + 1)^2}{16} + \frac{(y - 4)^2}{9} = 1\).
- (7-03) Sketch the graph of
*x*^{2}+ 16*y*^{2}+ 4*x*− 12 = 0. - (7-02) Write the standard equation for the parabola with vertex (2, 3) and directrix
*y*= −1. - (7-01) Find the angle between
*y*= −4*x*and 2*x*−*y*+ 3 = 0. - (2-04) Divide using synthetic division: (2
*x*^{3}+ 4*x*^{2}−*x*+ 3) ÷ (*x*+ 1).

Find the center, vertices, asymptotes, and foci of the following hyperbolas.

Find the standard equation of the hyperbola with the following properties.

Sketch the graph of the following hyperbolas.

Eccentricity

Problem Solving

Mixed Review

- The vertices are on the transverse axis. The covertices are on the conjugate axis.
- C(0, 0), V(0, ±8), A:
*y*= ±2*x*, F(0, ±\(4\sqrt{5}\)) - C(0, 0), V(\(±\sqrt{10}\), 0), A: \(y = ±\frac{2\sqrt{5}}{5} x\), F(±\(3\sqrt{2}\), 0)
- C(−6, 3), V(−9, 3) and (−3, 3), A: \(y = 3 ±\frac{2\sqrt{2}}{3} (x + 6)\), F(\(-6 ± \sqrt{17}\), 3)
- C(−1, 4), V(−1, −6) and (−1, 14), A: \(y = 4 ±\frac{10}{9} (x + 1)\), F(-1, 4 ± \(\sqrt{181}\))
- \(\frac{x^2}{4} - \frac{y^2}{12} = 1\)
- \(\frac{(y - 7)^2}{9} - \frac{(x - 3)^2}{16} = 1\)
- \(\frac{(y - 3)^2}{16} - \frac{(x - 1)^2}{36} = 1\)
- \(\frac{x^2}{9} - \frac{y^2}{7} = 1\)
- \(\frac{5}{3}\)
- \(\frac{x^2}{5625} - \frac{(y - 200)^2}{32000} = 1\)
- 5.099
- C(−1, 4), V(−5, 4) and (3, 4), CV(−1, 1) and (−1, 7), F(−1 ± \(\sqrt{7}\), 4)
- (
*x*− 2)^{2}= 16(*y*− 3) - 40.60°
- \(2x^2 + 2x - 3 + \frac{6}{x + 1}\)