10 5 Hyperbolas HYPERBOLA TERMS The Butterfly EQUATION

10. 5 - Hyperbolas

HYPERBOLA TERMS The “Butterfly” EQUATION FORM Conjugate axis CENTER VERTICES Co-vertex Vertex b Focus a c Transverse axis Vertex C=(h , k) Co-vertex CO-VERTICES (h, k ) (h ± a , k) (h, k ± b ) TRANSVERSE AXIS horizontal TRANSVERSE length 2 a CONJUGATE AXIS CONJUGATE length FOCI ASYMPTOTES vertical 2 b (h ± c , k)

HYPERBOLA TERMS The “Hourglass” Transverse axis Vertex EQUATION FORM CENTER VERTICES CO-VERTICES C=(h , k) Co-vertex c b TRANSVERSE AXIS TRANSVERSE length a Co-vertex Conjugate axis CONJUGATE AXIS CONJUGATE length Vertex FOCI ASYMPTOTES (h, k ) (h, k ± a ) (h ± b, k) vertical 2 a horizontal 2 b (h, k ± c )

CONVERTING to STANDARD FORM • • • 4 x² – 9 y² – 32 x – 18 y + 19 = 0 Groups the x terms and y terms 4 x² – 32 x – 9 y²– 18 y + 19 = 0 Complete the square 4(x² – 8 x) – 9(y² + 2 y) + 19 = 0 4(x² – 8 x + 16) – 9(y² + 2 y + 1) = -19 + 64 – 9 4(x – 4)² – 9(y + 1)² = 36 Divide to put in standard form 4(x – 4)²/36 – 9(y + 1)²/36 = 1

Example Graph 4 x 2 – 16 y 2 = 64 y 2 x 2 – =1 16 4 Rewrite the equation in standard form. The equation of the form y 2 x 2 – - 2 = 1, so the transverse axis is horizontal. a 2 b Since a 2 = 16 and b 2 = 4, a = 4 and b = 2. Step 1: Graph the vertices. Since the transverse axis is horizontal, the vertices lie on the x-axis. The coordinates are (±a, 0), or (± 4, 0). Step 2: Use the values a and b to draw the central “invisible” rectangle. The lengths of its sides are 2 a and 2 b, or 8 and 4.

Example Step 3: Graph 4 x 2 – 16 y 2 = 64. Draw the asymptotes. The equations of the asymptotes are y = ± b x or y = ± 1 x. The asymptotes contain the diagonals of the a central rectangle. Step 4: 2 Sketch the branches of the hyperbola through the vertices so they approach the asymptotes.

Graph and Label • b) Find coordinates of vertices, covertices, foci • Center = (-5, -2) • Butterfly shape since the x terms come first • Since a = 2 and b = 3 • Vertices are 2 points left and right from center (-5 ± 2, -2) • Co. Vertices are 3 points up and down (-5, -2 ± 3) • Now to find focus points • Use c² = a² + b² • So c² = 9 + 4 = 13 • c² = 13 and c = ±√ 13 • Focus points are √ 13 left and right from the center F(-5 ±√ 13 , -2) • • a) GRAPH Plot Center (-5, -2) a = 2 (go left and right) b = 3 (go up and down)

Graph and Label • b) Find coordinates of vertices, covertices, foci • Center = (-1, 3) • Hourglass shape since the y terms come first • Since a = 2 and b = 4 • Vertices are 2 points up and down from center (-1, 3 ± 2) • Covertices are 3 points left and right (-1 ± 4, 3) • Now to find focus points • Use c² = a² + b² • So c² = 4 + 16 = 20 • c² = 20 and c = ± 2√ 5 • Focus points are 2√ 5 up and down from the center F(-1, 3 ± 2√ 5) • • a) GRAPH Plot Center (-1, 3) a = 2 (go up and down) b = 4 (go left and right)

Write the equation of the hyperbola given… • • • Draw a graph with given info Use given info to get measurement Find the center first Center is in middle of vertices, so (h , k) = (0 , 2) A = distance from center to vertices, so a = 5 Also, the conjugate length = 2 b Since conjugate = 12 Then b = 6 Use standard form • • Need values for h, k, a and b We know a = 5 and b = 6 The center is (0, 2) Plug into formula A = (-5, 2) major conjugate vertices are at (-5, 2) and (5, 2) conjugate axis of length 12 B = (5, 2)

Write the equation of the hyperbola given… center is at (-3, 2) foci at (-3, 2± 13) and major axis is 10 • • Draw a graph with given info Use given info to get measurements Find the center first Center is in middle of vertices, so (h , k) = (0 , 2) A = distance from center to vertices, so a = 5 We still don’t have b …. • • • Use the formula c² = a² + b² Since a = 5 and c = 13 then…. b = 12 (pythagorean triplet) F(-3, 2+13) Use standard form Need values for h, k, a and b We know a = 5 and b = 12 and center is (0, 2) • Plug into formula • • F(-3, 2 -13)

Example y 2 x 2 Find the foci of the graph – = 1. 9 4 The equation is in the form y 2 x 2 – -2 a b 2 = 1, so the transverse axis is horizontal; a 2 = 9 and b 2 = 4. c 2 = a 2 + b 2 Use the Pythagorean Theorem. =9+4 Substitute 9 for a 2 and 4 for b 2. c= 13 3. 6 Find the square root of each side of the equation.

Example (continued) The foci (0, ±c) are approximately (0, – 3. 6) and (0, 3. 6). The vertices (0, ±b) are (0, – 2) and (0, 2). The asymptotes are the lines y = ± b 2 x , or y = ± x. a 3

Real Life Examples As a spacecraft approaches a planet, the gravitational pull of the planet changes the spacecraft’s path to a hyperbola that diverges from its asymptote. Find an equation that models the path of the spacecraft around the planet given that a = 300, 765 km and c = 424, 650 km. Assume that the center of the hyperbola is at the origin and that the transverse axis is horizontal. The equation will be in the form c 2 = a 2 + b 2 Use the Pythagorean Theorem. (424, 650)2 = (300, 765)2 + b 2 Substitute. 1. 803 1011 = 9. 046 1010 + b 2 y 2 x 2 – 2 = 1. a 2 b Use a calculator.

Real Life Examples (continued) b 2 = 1. 803 1011 – 9. 046 1010 Solve for b 2. = 8. 987 1010 x 2 9. 046 1010 – y 2 8. 987 1010 =1 Substitute a 2 and b 2. The path of the spacecraft around the planet can be modeled by x 2 9. 046 1010 – y 2 8. 987 1010 = 1.