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C O N I C S E C T I O N S

C O N I C S E C T I O N S. Part 4: Hyperbola. Hyperbola. Hyperbolas (opening left and right). Center: (h,k). y =. y =. Foci. Vertices. ( c , 0) . (– c , 0) . ( a , 0). (–a, 0). The foci of the hyperbola lie on the major axis, c units from the center, where c 2 = a 2 + b 2.

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C O N I C S E C T I O N S

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  1. C O N I CS E C T I O N S Part 4: Hyperbola

  2. Hyperbola

  3. Hyperbolas (opening left and right) Center: (h,k) y = y = Foci Vertices (c, 0) (–c, 0) (a, 0) (–a, 0) The foci of the hyperbola lie on the major axis, c units from the center, where c2 = a2+ b2 The Vertices are “a” distance from the center. (x - h)2 – (y – k)2 = 1 a2 b2 B-points (b, 0) (–b, 0) asymptotes The B-points are “b” distance from the center. The asymptotes can be found using the boxmethod, where the “a” and “b” points help form a box ….. The transverse axis is the line segment joining the vertices. The minor axis is vertical and acts as the line of reflection. It will contain the two “b” points. The major axis is horizontal and acts as the axis of symmetry. It will contain the Verticesand the Foci. Or you can use the linear equations.

  4. Hyperbolas (opening up and down) Center: (h,k) y = y = Foci Vertices (c, 0) (–c, 0) (a, 0) (–a, 0) The foci of the hyperbola lie on the major axis, c units from the center, where c2 = a2+ b2 The Vertices are “a” distance from the center. (y - k)2 – (x – h)2 = 1 a2 b2 b-points (b, 0) (–b, 0) asymptotes The b-points are “b” distance from the center. The asymptotes can be found using the boxmethod, where the “a” and “b” points help form a box ….. The transverse axis is the line segment joining the vertices. The major axis is vertical and acts as the axis of symmetry. It will contain the Verticesand the Foci. The minor axis is horizontal and acts as the line of reflection. It will contain the two “b” points. Or you can use the linear equations.

  5. Example: Write an equation of the hyperbola with foci (0, –6)and (0, 6)and vertices (0, –4)and(0, 4). Its center is (0, 0). vertical (0, 6) (0, 4) (–b, 0) (b, 0) a = 4,c = 6 c2=a2+b2 62 =42 +b2 36 = 16 + b2 20 = b2 The equation of the hyperbola: y2 – x2 = 1 16 20 (0, –4) (0, –6) (y – h)2 – (x – k)2 = 1 a2 b2

  6. vertical Draw the rectangle and asymptotes... Example: Graph y2 – x2 = 1 ; find foci and asymptotes 9 25 a = 3b = 5 c2 = a2 + b2 c2 = 9 + 25 = 34 c = (0, 3) (–5,0) (5, 0) Foci: Asymptotes: (0,–3)

  7. Get the equation in standard form (make it equal to 1): 4x2 – 16y2 = 6464 64 64 Example: Write the equation in standard form of4x2 – 16y2 = 64.Find the foci and vertices of the hyperbola. Simplify... x2 – y2 = 1 16 4 That means a = 4b = 2 Use c2 = a2 + b2 to findc. c2 =42 +22 c2 = 16 + 4 = 20 c = (0, 2) (–4,0) (4, 0) Vertices: Foci: (–c,0) (c, 0) (0,-2)

  8. Example: Graph (y – 2)2 – (x + 3)2 = 1 25 16 vertical Center: (–3, 2) a = 5 b = 4 (–3, 7) (–7, 2) (1, 2) To graph, start with the center… Move 5units up and down Move 4 units right and left (–3, –3) Draw the rectangle and asymptotes…

  9. Example: 9x2 – 4y2 + 18x + 16y – 43 = 0 9x2 + 18x –4y2 + 16y = 43 9(x2 + 2x ) – 4(y2 – 4y ) = 43 + 1 + 4 + 9 – 16 9(x + 1)2 – 4(y – 2)2 =36 (x + 1)2 – (y – 2)2 = 1 4 9 Center (–1, 2) a = 2b = 3 c2 = a2+ b2= 9 + 4 Foci: Asymptotes:

  10. Example: Write an equation in standard form for the hyperbola with vertices (–1, 1) and (7, 1) and foci (–2, 1) and (8, 1). Hyperbolas, write equations (3, 1) center: a = b = c = 4 5 c2=a2 + b2 25 = 16 + b2 b2 = 9 (x – 3)2 –(y – 1)2 = 1 16 9

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