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Hyperbola

Hyperbola. Conic Sections. The plane can intersect two nappes of the cone resulting in a hyperbola. Hyperbola. Hyperbola - Definition. A hyperbola is the set of all points in a plane such that the difference in the distances from two points (foci) is constant.

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Hyperbola

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  1. Hyperbola Conic Sections

  2. The plane can intersect two nappes of the cone resulting in a hyperbola. Hyperbola

  3. Hyperbola - Definition A hyperbola is the set of all points in a plane such that the difference in the distances from two points (foci) is constant. | d1 – d2 | is a constant value.

  4. Finding An Equation Hyperbola

  5. Hyperbola - Definition What is the constant value for the difference in the distance from the two foci? Let the two foci be (c, 0) and (-c, 0). The vertices are (a, 0) and (-a, 0). | d1 – d2| is the constant. If the length of d2 is subtracted from the left side of d1, what is the length which remains? | d1 – d2 | = 2a

  6. Hyperbola - Equation Find the equation by setting the difference in the distance from the two foci equal to 2a. | d1 – d2 | = 2a

  7. Hyperbola - Equation Simplify: Remove the absolute value by using + or -. Get one square root by itself and square both sides.

  8. Hyperbola - Equation Subtract y2 and square the binomials. Solve for the square root and square both sides.

  9. Hyperbola - Equation Square the binomials and simplify. Get x’s and y’s together on one side.

  10. Hyperbola - Equation Factor. Divide both sides by a2(c2 – a2)

  11. Hyperbola - Equation Let b2 = c2 – a2 where c2 = a2 + b2 If the graph is shifted over h units and up k units, the equation of the hyperbola is:

  12. Hyperbola - Equation where c2 = a2 + b2 Recognition:How do you tell a hyperbola from an ellipse? Answer:A hyperbola has a minus (-) between the terms while an ellipse has a plus (+).

  13. Graph - Example #1 Hyperbola

  14. Hyperbola - Graph Graph: Center: (-1, -2) The hyperbola opens in the “x” direction because “x” is positive. Transverse Axis: y = -2

  15. Hyperbola - Graph Graph: Vertices (2, -2) (-4, -2) Construct a rectangle by moving 4 unitsup and down from the vertices. Construct the diagonals of the rectangle.

  16. Hyperbola - Graph Graph: Draw the hyperbola touching the vertices and approaching the asymptotes. Where are the foci?

  17. Hyperbola - Graph Graph: The foci are 5 units from the center on the transverse axis. Foci: (-6, -2) (4, -2)

  18. Hyperbola - Graph Graph: Find the equation of the asymptote lines. 4 3 Use point-slope formy – y1 = m(x – x1) since the center is on both lines. -4 Slope = Asymptote Equations

  19. Graph - Example #2 Hyperbola

  20. Hyperbola - Graph Sketch the graph without a grapher: Recognition:How do you determine the type of conic section? Answer:The squared terms have opposite signs. Write the equation in hyperbolic form.

  21. Hyperbola - Graph Sketch the graph without a grapher:

  22. Hyperbola - Graph Sketch the graph without a grapher: Center: (-1, 2) Transverse Axis Direction: Up/Down Equation: x=-1 Vertices: Up/Down from the center or

  23. Hyperbola - Graph Sketch the graph without a grapher: Plot the rectangular points and draw the asymptotes. Sketch the hyperbola.

  24. Hyperbola - Graph Sketch the graph without a grapher: Plot the foci. Foci:

  25. Hyperbola - Graph Sketch the graph without a grapher: Equation of the asymptotes:

  26. Finding an EquationA problem for CSI! Hyperbola

  27. Hyperbola – Find an Equation The sound of a gunshot was recorded at one microphone 0.54 seconds before being recorded at a second microphone. If the two microphones are 2,000 ft apart.Provide a model for the possible locations of the gunshot. (The speed of sound is 1100 ft/sec.) The time between the shots can be used to calculate the difference in the distance from the two microphones. 1100 ft/sec * 0.54 sec = 594 ft. The constant difference in distance from the microphones is 594 ft. Since the difference is constant, the equation must be a hyperbola. The points on the hyperbola are possible positions for the gunshot.

  28. Hyperbola – Find an Equation Two microphones are stationed 2,000 ft apart. The difference in distance between the microphones is 594 ft. Let the center be at (0,0). The foci must be 2,000 ft apart. V The vertices are a possible position for the gunshot. The difference in the distance must be 594 feet between the vertices. Let the vertices be at (+z, 0). Assuming z>0, then(z-(-1000)) – (1000-z) = 594z+1000-1000+z = 5942z = 594 or z = 297.

  29. Hyperbola – Find an Equation V(-2970, 0) V(297, 0) V V Oops! We could have remembered the constant difference in distance is 2a! 2a = 594, a = 297. Start finding the model of the hyperbola. 2972 = 88209 The distance from the center to the foci (c) is 1000 ft. Find b.

  30. Hyperbola – Find an Equation V(294, 0) V(294, 0) V V The model is:

  31. Hyperbola – Find an Equation The gunshot was calculated to be at some point along the hyperbola.

  32. Conic Section Recogition

  33. Recognizing a Conic Section Parabola - One squared term. Solve for the term which is not squared. Complete the square on the squared term. Ellipse - Two squared terms. Both terms are the same “sign”. Circle - Two squared terms with the same coefficient. Hyperbola - Two squared terms with opposite “signs”.

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