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# Sullivan PreCalculus Section 9.4 The Hyperbola Objectives of this Section Find the Equation of a Hyperbola Graph Hyperbolas Discuss the Equation of a Hyperbola.

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Sullivan PreCalculus Section 9.4 The Hyperbola Objectives of this Section Find the Equation of a Hyperbola Graph Hyperbolas Discuss the Equation of a Hyperbola Find the Asymptotes of a Hyperbola Work with Hyperbolas with Center at (h,k)

A hyperbola is the collection of all points in the plane the difference of whose distances from two fixed points, called the foci, is a constant. F 2 : (c,0)F 1 : (-c,0) Transverse Axis

Equation of a Hyperbola: Center at (0,0); Foci at (c, 0) and (-c,0) where b 2 = c 2 - a 2 The transverse axis is the x - axis. The vertices are at (-a, 0) and (a, 0)

Find the equation of a hyperbola with center at the origin, one focus at (-5, 0), and a vertex at (4,0). Graph the equation. Since the given focus and vertex are on the x-axis, the transverse axis is the x-axis. The distance from the center to one of the foci is c = 5. The distance from the center to one of the vertices is a = 4. Use c and a to solve for b. b 2 = c 2 - a 2 b 2 = 5 2 - 4 2 = 25 - 16 = 9

So, the equation of the hyperbola is: F 2 : (5,0)F 1 : (-5,0) (-4,0)(4,0)

Discuss the equation: Since the equation is written in the desirable form, a 2 = 9 and b 2 = 7 Since b 2 = c 2 - a 2, it follows that c 2 = a 2 + b 2 or c 2 = 9 + 7 = 16. So, the foci are at (4,0) and (-4,0) The vertices are at (-3, 0) and (3, 0)

Discuss the equation: Since the x term is subtracted from the y term, the equation is that of a hyperbola with center at the origin and transverse axis along the y-axis The vertices are at (0,2) and (0,-2). Since b 2 = c 2 - a 2, it follows that c 2 = a 2 + b 2 or c 2 = 4 + 1 = 5. So, the foci are at and

The hyperbola has the two oblique asymptotes: Discuss the equation of the hyperbola Begin by dividing both sides of the equation by 16 to put the equation in the proper form.

The center of the hyperbola is the origin. Since the x term comes first, the transverse axis is the x-axis. The vertices are at (1,0) and (-1,0). Since b 2 = c 2 - a 2, it follows that c 2 = a 2 + b 2 or c 2 = 1 + 4 = 5. So, the foci are at and

The asymptotes have the equation: To graph the hyperbola, form the rectangle containing the points (a,0), (-a,0), (0,b), and (0,-b). The extensions of the diagonals of this rectangle are the asymptotes.

(1,0) (-1,0) (0,-2) (0,2) Now, graph the hyperbola using these guides

If an ellipse with center at the origin and major axis coinciding with a coordinate axis is shifted horizontally h units and vertically k units, the resulting ellipse is centered at (h,k) and has the equation: Horizontal Transverse Axis Vertical Transverse Axis

Find the equation of a hyperbola with center at (2, -3), one focus at (5, -3), and one vertex at (3, -3). The center is at (h,k) = (2, -3). So h = 2 and k = -3 The center, focus, and vertex all lie on the line y = -3, so the major axis is parallel to the x-axis and the hyperbola has a horizontal transverse axis and will have an equation in the form:

The distance from the center to the vertex is a = 1. The distance from the center to the focus is c = 3. To solve for b, b 2 = c 2 - a 2 b 2 = 3 2 - 1 2 = 9 - 1 = 8 So, the equation of the ellipse is:

The equations of the asymptotes can be found by shifting the equations for the asymptotes h units in the horizontal direction and k units in the vertical direction yielding: For our example:

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