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HYPERBOLA. PARTS OF A HYPERBOLA center Focus 2 Focus 1 conjugate axis vertices The dashed lines are asymptotes for the graphs transverse axis.

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Presentation on theme: "HYPERBOLA. PARTS OF A HYPERBOLA center Focus 2 Focus 1 conjugate axis vertices The dashed lines are asymptotes for the graphs transverse axis."— Presentation transcript:

1 HYPERBOLA

2 PARTS OF A HYPERBOLA center Focus 2 Focus 1 conjugate axis vertices The dashed lines are asymptotes for the graphs transverse axis

3 A hyperbola is the collection of points in the plane whose difference of distances from two fixed points, called the foci, is a constant. P (x,y) |dPF 1 – dPF 2 |= 2a

4 Curves open sideways: Horizontal Transverse axis The x-axis contains: - 2 vertices, (-a, 0) and (a, 0) - 2 foci at (-c, 0), (c, 0) - has length = 2a * the conjugate axis (y-axis) has length = 2b F(-c, 0) F(c, 0) For every point P on the hyperbola we have |dPF 1 – dPF 2 |= 2a

5 Curves open up/down:Vertical Transverse axis. The y-axis contains: - 2 vertices, (0, -b) and (0, b) - 2 foci at (0, -c), (0, c) - has length = 2b * the conjugate axis (x-axis) has length = 2a (0, -b) F(0, -c) F(0, c) For every point P on the hyperbola we have |dPF 1 – dPF 2 |= 2b (0, b)

6 HYPERBOLA with any center:

7 Pythagorean Property

8 a a c b b Graphing Hyperbolas Step 1: Make a rectangle measuring 2a by 2b and draw the diagonals. These are the asymptotes. Step 2: Plot the vertices on the transverse axis and draw the arches of the hyperbola

9 Example 1 Graph the hyperbola 4x 2 – 16y 2 = 64, and find the focal distance. 4x 2 – 16y 2 = 64 64 64 64 (–4,0)(4, 0) (0, 2) (0,-2) That means a = 4 and b = 2 c2 = a2+ b2c2 = a2+ b2 c 2 = 16 + 4 c 2 = 20 c = ±√20 Focal distance is 2√20

10 Example 2: Write an equation of the hyperbola centered at (0,0) whose foci are (0, –6) and (0, 6) and whose vertices are (0, –4) and (0, 4). b= 4 and c = 6 c 2 = a 2 + b 2 6 2 = a 2 + 4 2 36 = 16 + a 2 20 = a 2 (0, 4) (0, –4) (0, 6) (0, –6)

11 Example 3: Determine the inequality of the shaded hyperbolic region with vertex (0,12) and focus (0,15). b= 12 and c = 15 c 2 = a 2 + b 2 225 = a 2 + 144 81 = a 2

12 Example 4: Determine the inequality of the hyperbola representing the shaded region. Slope of asymptote is y = bx a Slope of asymptote: -4-0 = 4 -3-0 3 so b = 4 a 3 b = 4 6 3 b = 8

13 Example 5: Determine the intersection point(s), if any, of the line 2x – y + 1 = 0 and the hyperbola


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