1 / 14

10.5 - Hyperbolas

10.5 - Hyperbolas. Vertex. HYPERBOLA TERMS. The “Butterfly”. (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). Vertex. HYPERBOLA TERMS.

margaret-walton
Download Presentation

10.5 - Hyperbolas

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 10.5 - Hyperbolas

  2. Vertex HYPERBOLA TERMS The “Butterfly” (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)

  3. Vertex HYPERBOLA TERMS The “Hourglass” 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 )

  4. 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

  5. 4x2 – 16y2 = 64 – = 1 Rewrite the equation in standard form. The equation of the form – - = 1, so the transverse axis is horizontal. Since a2 = 16 and b2 = 4, a = 4 and b = 2. y2 y2 x2 x2 16 a2 4 b2 Example Graph 4x2 – 16y2 = 64. 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.

  6. b a 1 2 Step 4: Sketch the branches of the hyperbola through the vertices so they approach the asymptotes. 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.

  7. Graph and Label • a) GRAPH • Plot Center (-5,-2) • a = 2 (go left and right) • b = 3 (go up and down) • 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)

  8. Graph and Label • a) GRAPH • Plot Center (-1,3) • a = 2 (go up and down) • b = 4 (go left and right) • 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)

  9. 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

  10. Write the equation of the hyperbola given…center is at (-3,2) foci at (-3, 2±13) and major axis is 10 F(-3,2+13) • 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)

  11. y2 The equation is in the form – - = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4. y2 x2 b2 9 4 c2 = a2 + b2Use the Pythagorean Theorem. = 9 + 4Substitute 9 for a2 and 4 for b2. x2 a2 c = 13 3.6 Find the square root of each side of the equation. Example Find the foci of the graph – = 1.

  12. b a 2 3 The asymptotes are the lines y = ± x , or y = ± x. 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).

  13. 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. x2 a2 c2 = a2 + b2Use the Pythagorean Theorem. y2 b2 (424,650)2 = (300,765)2 + b2Substitute. 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. 1.803  1011 = 9.046  1010 + b2Use a calculator.

  14. b2 = 1.803  1011 – 9.046  1010Solve for b2. = 8.987  1010 – = 1 Substitute a2 and b2. – = 1. The path of the spacecraft around the planet can be modeled by x2 x2 y2 y2 9.046  1010 9.046  1010 8.987  1010 8.987  1010 Real Life Examples (continued)

More Related