10.5 - Hyperbolas

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

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

CONVERTING to STANDARD FORM 4x² – 9y² – 32x – 18y + 19 = 0 Groups the x terms and y terms 4x² – 32x – 9y²– 18y + 19 = 0 Complete the square 4(x² – 8x) – 9(y² + 2y) + 19 = 0 4(x² – 8x + 16) – 9(y² + 2y + 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 4x2 – 16y2 = 64. 4x2 – 16y2 = 64 – = 1 Rewrite the equation in standard form. x 2 16 y 2 4 Since a2 = 16 and b2 = 4, a = 4 and b = 2. The equation of the form – - = 1, so the transverse axis is horizontal. x 2 a 2 y 2 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 2a and 2b, or 8 and 4.

Example Graph 4x2 – 16y2 = 64. Step 3: Draw the asymptotes. The equations of the asymptotes are y = ± x or y = ± x . The asymptotes contain the diagonals of the central rectangle. b a 1 2 Step 4: 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) CoVertices 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… vertices are at (-5,2) and (5,2) conjugate axis of length 12 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 = 2b 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 B = (5,2) A = (-5,2) conjugate major

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) 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) F(-3,2-13)

Example Find the foci of the graph – = 1. y 2 4 Find the foci of the graph – = 1. x 2 9 The equation is in the form – - = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4. x 2 a2 y 2 b2 c2 = a2 + b2 Use the Pythagorean Theorem. = 9 + 4 Substitute 9 for a2 and 4 for b2. 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 = ± x , or y = ± x. b a 2 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 – = 1. x 2 a2 y 2 b2 c2 = a2 + b2 Use the Pythagorean Theorem. (424,650)2 = (300,765)2 + b2 Substitute. 1.803  1011 = 9.046  1010 + b2 Use a calculator.

Real Life Examples b2 = 1.803  1011 – 9.046  1010 Solve for b2. (continued) b2 = 1.803  1011 – 9.046  1010 Solve for b2. = 8.987  1010 – = 1 Substitute a2 and b2. x 2 9.046  1010 y 2 8.987  1010 The path of the spacecraft around the planet can be modeled by – = 1. x 2 9.046  1010 y 2 8.987  1010