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- 1. CONICS HYPERBOLA
- 2. Hyperbola • A conic section that can be thought of as an inside-out ellipse. • A hyperbola is the locus of points where the difference in the distance to two fixed points (called the foci) is constant.
- 3. • A hyperbola is the set of all points (x, y) in the plane the difference of whose distances from two fixed points is some constant. The two fixed points are called the foci. • A hyperbola comprises two disconnected curves called its arms or branches which separate the foci.
- 4. Elements of the Hyperbola
- 5. • Foci The foci are the fixed points of the hyperbola. They are denoted by F and F'. • Transverse Axis or real axis The tranverse axis is the line segment between the foci. • Conjugate Axis or imaginary axis The conjuagate axis is the perpendicular bisector of the line segment (transverse axis). • Center The center is the point of intersection of the axes and is also the center of symmetry of the hyperbola.
- 6. • Vertices The points A and A' are the points of intersection of the hyperbola with the transverse axis. • Focal Radii The focal radii are the line segments that join a point on the hyperbola with the foci: PF and PF'. • Focal Length The focal length is the line segment , which has a length of 2c.
- 7. • Semi-Major Axis The semi-major axis is the line segment that runs from the center to a vertex of the hyperbola. Its length is a. • Semi-Minor Axis The semi-minor axis is a line segment which is perpendicular to the semi-major axis. Its length is b. • Axes of Symmetry The axes of symmetry are the lines that coincide with the transversal and conjugate axis.
- 8. • Asymptotes The asymptotes are the lines with the equations: • The Relationship Between the Semi axes: • Eccentricity of a Hyperbola:
- 9. Hyperbola is actually two separate curves in mirror image. On the diagram you can see: •a directrix and a focus (one on each side) •an axis of symmetry or transverse axis (that goes through each focus, at right angles to the directrix) •two vertices (where each curve makes its sharpest turn)
- 10. Equation • By placing a hyperbola on an x-y graph (centered over the x-axis and y-axis), the equation of the curve is: x2/a2 - y2/b2 = 1
- 11. Also: One vertex is at (a, 0), and the other is at (-a, 0) The asymptotes are the straight lines: •y = (b/a)x •y = -(b/a)x And the equation is also similar to the equation of the ellipse: x2/a2+ y2/b2 = 1, except for a "-" instead of a "+")
- 12. When the transverse axis is horizontal (in other words, when the center, foci, and vertices line up side by side, parallel to the x-axis), then the a2 goes with the x part of the hyperbola's equation, and the y part is subtracted.
- 13. When the transverse axis is vertical (in other words, when the center, foci, and vertices line up above and below each other, parallel to the y-axis), then the a2 goes with the y part of the hyperbola's equation, and the x part is subtracted.
- 14. Asymptotes The graph of a hyperbola gets fairly flat and straight when it gets far away from its center. If you "zoom out" from the graph, it will look very much like an "X", with a little curviness near the middle. These "nearly straight" parts get very close to what are called the "asymptotes" of the hyperbola. For a hyperbola centered at (h, k) and having fixed values a and b, the asymptotes are given by the following equations:
- 15. Hyperbolas’ Graphs Asymptotes’ Equations
- 16. Note that the only difference in the asymptote equations is in the slopes of the straight lines: If a2is the denominator for the x part of the hyperbola's equation, then a is still in the denominator in the slope of the asymptotes' equations.
- 17. If a2 goes with the y part of the hyperbola's equation, then a goes in the numerator of the slope in the asymptotes' equations.
- 18. Eccentricity • The measure of the amount of curvature is the "eccentricity" e, where e = c/a. Since the foci are further from the center of an hyperbola than are the vertices (so c > a for hyperbolas), then e > 1. Bigger values of e correspond to the "straighter" types of hyperbolas, while values closer to 1 correspond to hyperbolas whose graphs curve quickly away from their centers.
- 19. On this diagram: • P is a point on the curve, • F is the focus and • N is the point on the directrix so that PN is perpendicular to the directrix. The ratio PF/PN is the eccentricity of the hyperbola (for a hyperbola the eccentricity is always greater than 1). It can also given by the formula: e =
- 20. Hyperbolas can be fairly "straight" or else pretty "bendy": Hyperbola with an eccentricity of about 1.05 Hyperbola with an eccentricity of about 7.6
- 21. Latus Rectum The Latus Rectum is the line through the focus and parallel to the directrix. The length of the Latus Rectum is 2b2/a.
- 22. Hyperbola on the Horizontal Axis Center at the Origin Equation: x2/a2 - y2/b2 = 1 C (0,0) V (± a,0) F (±c,0) Length of LR = 2b2/a Ends of LR (c,±b2/a) and (-c,±b2/a) Ends of Conjugate Axis (0,±b) Asymptotes y = ± (b/a)x e = c/a c2 = a2+b2
- 23. Hyperbola on the Horizontal Axis Center at (h,k) Equation: (x-h)2/a2 - (y-k)2/b2 = 1 C (h,k) V (h± a,k) F (h±c,k) Length of LR = 2b2/a Ends of LR (h-c,k±b2/a) and (h-c,k±b2/a) Ends of Conjugate Axis (h,k±b) Asymptotes y = ± (b/a)x e = c/a c2 = a2+b2
- 24. Hyperbola on the Vertical Axis Center on the Origin Equation: y2/a2 - x2/b2 = 1 C (0,0) V (0,± a) F (0,±c) Length of LR = 2b2/a Ends of LR (±b2/a,c) and (±b2/a,-c) Ends of Conjugate Axis (±b,0) Asymptotes y = ± (a/b)x e = c/a c2 = a2+b2
- 25. Hyperbola on the Vertical Axis Center at (h,k) Equation: (y-k)2/a2 - (x-h)2/b2 = 1 C (h,k) V (h,k± a) F (h,k±c) Length of LR = 2b2/a Ends of LR (h±b2/a,k+c) and (h±b2/a,k-c) Ends of Conjugate Axis (h±b,k) Asymptotes y = ± (a/b)x e = c/a c2 = a2+b2
- 26. Exercises 1.) Find the center, vertices, foci, eccentricity, and asymptotes of the hyperbola with the given equation, and sketch: . 2.) Give the center, vertices, foci, and asymptotes for the hyperbola with equation: .
- 27. 3.) Find the center, vertices, and asymptotes of the hyperbola with equation 4x2 – 5y2 + 40x – 30y – 45 = 0. 4.) Find an equation for the hyperbola with center (2, 3), vertex (0, 3), and focus (5, 3). 5.) Find an equation for the hyperbola with center (0, 0), vertex (0, 5), and asymptotes y = ± (5/3)x.
- 28. 6.) Find an equation of the hyperbola with xintercepts at x = –5 and x = 3, and foci at (–6, 0) and (4, 0). 7.) Find an equation for the hyperbola with vertices at (–2, 15) and (–2, –1), and having eccentricity e = 17/8. 8.) Give the equation of the hyperbola with vertices at (-7,0) and (7,0) and conjugate axis length of 10.
- 29. Answers: 1.) c2 = a2 + b2 c2 = 144 + 25 = 169 c = 13 e = c/a = 13/5 Since x2 = (x – 0)2 and y2 = (y – 0)2, then the center is at (h, k) = (0, 0). F (0,-13) and (0,13) V (0,5) and (0,-5)
- 30. center (0, 0), vertices (0, –5) and (0, 5), foci (0, –13) and (0, 13), eccentricity , and asymptotes
- 31. 2.) a2 = 16 and b2 = 9, so a = 4 and b = 3. C = (h, k) = (–3, 2) c2 = 9 + 16 = 25 c=5 F = (h-c, k) = (-3-5, 2) = (-8, 2) Asymptotes = (h+c, k) = (-3+5, 2) = (2, 2) V = (h-a, k) = (-3-4, 2) = (-7, 2) y – 2 = ± (3/4)(x + 3). = (h+a, k) = (-3+4, 2) = (1,2)
- 32. 3.) 4x2 – 5y2 + 40x – 30y – 45 = 0 4x2 + 40x – 5y2 – 30y = 45 4(x2 + 10x ) – 5(y2 + 6y ) = 45 + 4( ) – 5( ) 4(x2 + 10x + 25) – 5(y2 + 6y + 9) = 45 + 4(25) – 5(9) 4(x + 5)2 – 5(y + 3)2 = 45 + 100 – 45
- 33. C = (h, k) = (–5, –3) a2 = 25 b2 = 20 a=5 b = 2sqrt[5] c2 = 25 + 20 = 45 c =sqrt[45] = 3sqrt[5] Asymptotes = m = ± (2/5)sqrt[5] center (–5, –3), vertices (–10, –3) and (0, –3), foci and , and asymptotes
- 34. 4.) C (2, 3), V (0, 3), and F (5, 3) V (h+a, k) 2-a = 0 -a = -2 a=2 Eq. F (h+c, k) 2+c = 5 c = 5-2 c=3
- 35. 5.) C (0, 0), V (0, 5), and asymptotes y = ± (5/3)x V (h, k+a) k+a = 5 0+a = 5 a=5 a2 = 25 Eq. asymptotes = a/b = 5/3 = 5/b b=3 b2 = 9
- 36. 6.) x-intercepts at x = –5 and x = 3, and foci at (–6, 0) and (4, 0) C (h, k) = (–1, 0) The foci are 5 units to either side of the center, so c = 5 and c2 = 25 The center lies on the x-axis, so the two xintercepts must then also be the hyperbola's vertices. Since the intercepts are 4 units to either side of the center, then a = 4 and a2 = 16. a2 +b2 = c2 b2 = 25 – 16 = 9 Eq.
- 37. 7.) V (–2, 15) and (–2, –1), and e = 17/8 (h, k) = (–2, 7) The vertices are 8 units above and below the center, so a = 8 and a2 = 64. e = c/a a2 + b2 = c2 =17/8 b2 = 289 – 64 = 225 c = 17 c2 = 289 Eq.
- 38. 8.) V (-7,0) (7,0) length of conjugate axis = 10 The hyperbola has its center at the origin because the center is halfway between the vertices and the origin is halfway between the vertices. The equation of a hyperbola which opens right and left is where a = half the length of the transverse axis (the transverse axis is the line between the vertices) and b = half the length of the conjugate axis (the conjugate axis is a line perpendicular to the transverse axis at the center, half of which is above the center and half below.
- 39. So since the vertices are (-7,0) and (7,0) the major axis is 14 and half that is 7, so a = 7 Half the conjugate axis is half of 10, or 5. So b = 5. So the equation is: or
- 40. References • http://mathworld.wolfram.com/Hyperbola.html • http://www.purplemath.com/modules/hyperbola 1.htm • http://www.mathsisfun.com/geometry/hyperbola .html • http://www.mathwords.com/h/hyperbola.htm
- 41. • http://www.mathportal.org/analyticgeometry/conic-sections/hyperbola.php • http://www.ditutor.com/conics/hyperbola.html • http://www.clausentech.com/lchs/dclausen/algeb ra2/hyperbolas.htm • http://dictionary.reference.com/browse/conjugat e+axis • http://www.barstow.edu/lrc/tutorserv/113ellhy.p df
- 42. • http://www.algebra.com/algebra/homework/Qua dratic-relations-and-conic-sections/Quadraticrelations-and-conicsections.faq.question.80329.html
- 43. GROUP 3 Jasper Neil Caparros Alekcis Gayares Gabriel Antonio Don Joseph Andreus Patiño Charmaine Lacsado Grace Penales Alyssa Villanueva BSFT2-1

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