Section 8.3 Hyperbolas

Definition: A hyperbola is the set of all points, the difference of whose distances from two fixed points is constant. Each fixed point is called a focus (plural = foci ).

The focal axis is the line passing through the foci.

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Basic “vertical” hyperbola:

y2 x 2 a (0,c ) Equation: − = 1 y= − x a2 b 2 b a y= x b a a Asymptotes: y= ± x b

Foci: (0,±c ) , where c2= a 2 + b 2

Vertices: (0,±a ) (0,−c ) -a c Eccentricity: ( > )1 a

Basic “horizontal” hyperbola: b b y= − x y= x a a x2 y 2 Equation: − = 1 a2 b 2

b Asymptotes: y= ± x a

± 2= 2 + 2 -a a Foci: (c ,0) , where c a b (c ,0) (c ,0)

Vertices: (±a ,0)

c Eccentricity: ( > )1 a

Note: The transverse axis is the line segment joining the two vertices. The conjugate axis is the line segment perpendicular to the transverse axis, passing through the center and extending a distance b on either side of the center. (These terms will make more sense after we do the graphing examples.)

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Graphing hyperbolas:

To graph a hyperbola with center at the origin:

x2 y 2 y2 x 2 • Rearrange into the form − = 1 or − = 1. a2 b 2 a2 b 2

• Decide if it’s a “horizontal” or “vertical” hyperbola. o if x2 comes first, it’s horizontal (vertices are on x-axis). o If y2 comes first, it’s vertical (vertices are on y-axis). • Use the square root of the number under x2 to determine how far to measure in x- direction.

• Use the square root of the number under y2 to determine how far to measure in y- direction.

• Draw a box with these measurements.

• Draw diagonals through the box. These are the asymptotes. Use the dimensions of the box to determine the slope and write the equations of the asymptotes.

• Put the vertices at the edge of the box on the correct axis. Then draw a hyperbola, making sure it approaches the asymptotes smoothly.

2 2 2 2 2 • c= a + b where a and b are the denominators.

• The foci are located c units from the center, on the same axis as the vertices.

When graphing hyperbolas, you will need to find the orientation, center, values for a, b and c, lengths of transverse and converse axes, vertices, foci, equations of the asymptotes, and eccentricity.

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x2 y2 Example 1: Find all relevant information and graph − = 1. 36 4

Vertices:

Foci:

Eccentricity:

Transverse Axis:

Length of transverse axis:

Conjugate axis:

Length of conjugate axis:

Slant Asymptotes:

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y2 x 2 Example 2: Find all relevant information and graph − = 1 . 4 9

Vertices:

Foci:

Eccentricity:

Transverse Axis:

Length of transverse axis:

Conjugate axis:

Length of conjugate axis:

Slant Asymptotes:

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The equation of a hyperbola with center not at the origin: Center: (h, k)

(xh− )(2 yk − ) 2 (yk− )(2 xh − ) 2 − = 1 or − = 1 a2 b 2 a2 b 2

To graph a hyperbola with center not at the origin:

• Rearrange (complete the square if necessary) to look like (xh− )(2 yk − ) 2 (yk− )(2 xh − ) 2 − = 1 or − = 1. a2 b 2 a2 b 2 • Start at the center (h , k ) and then graph it as before. • To write down the equations of the asymptotes, start with the equations of the asymptotes for the similar hyperbola with center at the origin. Then replace x with x− h and replace y with y− k .

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Example 3: Write the equation in standard form, find all relevant information and graph

9x 2 −16 y 2 −18 x + 96 y = 279 .

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Example 4: Write an equation of the hyperbola with center at (-2, 3), one vertex is at (-2, -2) and eccentricity is 2.

Example 5: Write an equation of the hyperbola if the vertices are (4, 0) and (4, 8) and the asymptotes have slopes ±1.

10 Math 1330 - Chapter 8 Systems: Identify Equations, Point of Intersection of Equations

Classification of Second Degree Equations

When you write a conic section in its general form, you have an equation of the form Ax2+ Bxy + Cy 2 + Dx ++= Ey F 0. (All of the equations we have seen so far have a value for B that is 0.)

We graphed the following examples in the past sections:

5x 2 + 5y 2 − 20 x +10 y = 20 (a circle)

y 2 − 6y = 8x + 7 (a parabola)

4x2 − 8x + 9y2 − 54 y −= 49 (an ellipse)

9x 2 −16 y 2 −18 x + 96 y = 279 (a hyperbola)

With only minimal work, you can determine if an equation in this form is a circle, an ellipse, a parabola or a hyperbola.

Identify each conic section from its equation:

(x − 2)2 (y + 2)2 a) 12 x = y 2 b) − = 1 9 16

(x + 4)2 (y −1)2 (x + 4)2 (y −1)2 c) + = 1 d) + = 1 4 9 4 4

1 Classification of Second Degree Equations

When you write a conic section in its general form, you have an equation of the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 :

If A, B and C are not all 0, and if the graph is not degenerate, then:

• The graph is a circle if B 2 − 4AC < 0 and A = C . • The graph is an ellipse if B 2 − 4AC < 0 and A ≠ C . • The graph is a parabola if B 2 − 4AC = 0 . • The graph is a hyperbola if B 2 − 4AC > 0 .

Remember, if there is no “xy” term, then B = 0.

Example: Identify each conic. (Note, none of these are degenerates conics.) a. 6x 2 − 4xy + 3y 2 + 5x − 7y + 3 = 0

b. 2x 2 − 8y 2 − 6x −16 y − 25 = 0

c. − 3x 2 + 5x −12 y − 7 = 0

d. 4x 2 + 4y 2 − 24 x −16 y − 72 = 0

2 Sometimes equations that look like they should be conic sections do not behave very well.

Example 1 : Graph (x − 3)2 + (y +1)2 = 0

Example 2: Graph 9x 2 − 4y 2 = 0

3 Example 3: Graph 2y + (x + 2)2 = 2y

Example 4: Graph 2x 2 + 3y 2 = − 1

These are all examples of degenerate conic sections . Instead of getting the graphs you expect, you have a point (Example 1), two lines (Example 2) and a single line (Example 3) and no graph at all (Example 4). You will not see these very often, but you should be aware of them.

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Systems of Second Degree Equations

When we graph two conic sections or a conic section and a line on the same coordinate plane, their graphs may contain points of intersection. The graph below shows a hyperbola and a line and contains two points of intersection.

We want to be able to find the points of intersection. To do this, we will solve a system of equations, but now one or both of the equations will be second degree equations. Determining the points of intersection graphically is difficult, so we will do these algebraically.

Example 5: Determine the number of points of intersection for the system.

x2 y 2 − = 1 9 9 y2 = 4 x

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Example 6: Solve the system of equations:

x 2 + y 2 = 4

4x 2 − y 2 = 1

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Example 7: Solve the system of equations: f x)( = x 2 − 4x +11 g x)( = 5x − 3

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Example 8: Solve the system of equations:

x 2 + y 2 = 9

()x + y 2 = 9

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Example 9: Graph each and determine the number of points of intersection. x2+ y 2 = 16 x2 y 2 + = 1 9 4

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