Report

HYPERBOLA PARTS OF A HYPERBOLA conjugate axis The dashed lines are asymptotes for the graphs transverse axis center Focus 1 Focus 2 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) |dPF1 – dPF2|= 2a Curves open sideways: Horizontal Transverse axis x2 y2 2 1 2 a b 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 For every point P on the hyperbola we have |dPF1 – dPF2|= 2a F(-c, 0) F(c, 0) Curves open up/down:Vertical Transverse axis. The y-axis contains: x2 y2 2 1 2 a b - 2 vertices, (0, -b) and (0, b) - 2 foci at (0, -c), (0, c) F(0, c) (0, b) - has length = 2b * the conjugate axis (x-axis) has length = 2a For every point P on the hyperbola we have |dPF1 – dPF2|= 2b (0, -b) F(0, -c) HYPERBOLA with any center: ( x h) ( y k) 1 2 2 a b 2 2 ( x h) ( y k) 1 2 2 a b 2 2 Pythagorean Property 2 2 2 2 x y 1 2 2 a b or x y 1 2 2 a b 2 2 2 where c a b Graphing Hyperbolas 2 2 2 x y 1 4 16 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 2 x y 2 1 2 a b b a b c a Example 1 2 2 Graph the hyperbola 4x – 16y = 64, and find the focal distance. 2 2 4x – 16y = 64 64 64 64 x2 y 2 That means a = 4 and b = 2 1 16 4 2 (0, 2) (–4,0) (4, 0) (0,-2) 2 2 c =a +b 2 c = 16 + 4 2 c = 20 c = ±√20 Focal distance is 2√20 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). (0, 6) (0, 4) (0, –4) (0, –6) b= 4 and c = 6 2 2 2 c =a +b 2 2 2 6 =a +4 2 36 = 16 + a 2 20 = a 2 2 x y 1 20 16 Example 3: Determine the inequality of the shaded hyperbolic region with vertex (0,12) and focus (0,15). b= 12 and c = 15 2 2 2 c =a +b 2 225 = a + 144 2 81 = a 2 2 x y 1 81 144 2 2 x y 1 81 144 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=8 b=4 6 3 2 2 x2 y2 x y 1 1 36 64 36 64 Example 5: Determine the intersection point(s), if any, of the line 2x – y + 1 = 0 and the hyperbola 2 2 x y 1 8 13