Precalculus Unit 5 Hyperbolas

1. Precalculus Unit 5 Hyperbolas

2. Hyperbolas A hyperbola is a set of points in a plane the difference of whose distances from two fixed points, called foci, is a constant. For any point P that is on the hyperbola, d2 – d1 is always the same. P d2 d1 F1 F2 In this example, the origin is the center of the hyperbola. It is midway between the foci.

3. Hyperbolas A line through the foci intersects the hyperbola at two points, called the vertices. The segment connecting the vertices is called the transverseaxis of the hyperbola. F C F V V The center of the hyperbola is located at the midpoint of the transverse axis. The distance from the center to a vertex is a. As x and y get larger the branches of the hyperbola approach a pair of intersecting lines called the asymptotes of the hyperbola. These asymptotes pass through the center of the hyperbola.

4. Hyperbolas The figure at the left is an example of a hyperbola whose branches open up and down instead of right and left. F V Since the transverse axis is vertical, this type of hyperbola is often referred to as a vertical hyperbola. C V F When the transverse axis is horizontal, the hyperbola is referred to as a horizontalhyperbola.

5. Standard Form Equation of a Hyperbola (x – h)2(y – k)2 (y – k)2(x – h)2 – = 1 – = 1 a2 b2 b2 a2 Horizontal Hyperbola Vertical Hyperbola The center of a hyperbola is at the point (h, k) in either form For either hyperbola, c2 = a2 + b2 Where c is the distance from the center to a focus point. The equations of the asymptotes are ba - ba and y = (x – h) + k y = (x – h) + k

6. Finding the Center, Vertices and Foci Ex : x2 y2 16 7 – = 1 Center: (0, 0) The x-term comes first in the subtraction so this is a horizontal hyperbola a 2 =16 so a = 4. The vertices are to the right and left of center x 4 spaces. Vertices: (4, 0) and (-4, 0) From the pythagorean relation c2 = a2 + b2 c2 = 16 + 7 = 23 c = Ö23 Foci: (Ö23, 0) and (-Ö23, 0)

7. Graphing a Hyperbola Graph: x2 y2 4 9 – = 1 Center: (0, 0) The x-term comes first in the subtraction so this is a horizontal hyperbola From the center locate the points that are two spaces to the right and two spaces to the left From the center locate the points that are up three spaces and down three spaces Draw a dotted rectangle through the four points you have found. Draw the asymptotes as dotted lines that pass diagonally through the rectangle. c2 = 9 + 4 = 13 c = Ö13 = 3.61 Draw the hyperbola. Foci: (3.61, 0) and (-3.61, 0) Vertices: (2, 0) and (-2, 0)

8. Graphing a Hyperbola Graph: (x + 2)2 (y – 1)2 9 25 – = 1 Horizontal hyperbola Center: (-2, 1) Vertices: (-5, 1) and (1, 1) c2 = 9 + 25 = 34 c = Ö34 = 5.83 Foci: (-7.83, 1) and (3.83, 1) 53 Asymptotes: y = (x + 2) + 1 53 - y = (x + 2) + 1

9. Converting an Equation Graph: 9y2 – 4x2 – 18y + 24x – 63 = 0 9(y2 – 2y + ___) – 4(x2 – 6x + ___) = 63 + ___ – ___ 1 9 9 36 9(y – 1)2 – 4(x – 3)2 = 36 (y – 1)2 (x – 3)2 4 9 – = 1 The hyperbola is vertical Center: (3, 1) c2 = 9 + 4 = 13 c = Ö13 = 3.61 Foci: (3, 4.61) and (3, -2.61) 23 Asymptotes: y = (x – 3) + 1 23 - y = (x – 3) + 1

10. Finding an Equation Find the standard form of the equation of a hyperbola given: Foci: (-7, 0) and (7, 0) Vertices: (-5, 0) and (5, 0) 8 Horizontal hyperbola Center: (0, 0) a2 = 25 and c2 = 49 F V C V F 10 c2 = a2 + b2 49 = 25 + b2 b2 = 24 (x – h)2(y – k)2 – = 1 a2 b2 x2 y2 – = 1 25 24

11. Finding an Equation Find the standard form equation of the hyperbola that is graphed at the right Vertical hyperbola (y – k)2(x – h)2 – = 1 b2 a2 Center: (-1, -2) a = 3 and b = 5 (y + 2)2(x + 1)2 – = 1 25 9

12. General Effects of the Parameters A and C When A ≠ C, and A x C < 0, the resulting conic is an hyperbola. 3.5.18

13. Definition of Eccentricity of an Ellipse The eccentricity of an ellipse is

14. Homework: Page 663 2-16 even, 24 – 40 even, 47, 52