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The Hyperbola. Definition of a hyperbola. The set of all points in a plane such that the difference of the distances to 2 fixed points (foci) is constant. Standard form of a hyperbola centered at the origin: -Opens left & right -Opens up & down.
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Definition of a hyperbola • The set of all points in a plane such that the difference of the distances to 2 fixed points (foci) is constant. • Standard form of a hyperbola centered at the origin: -Opens left & right -Opens up & down
What are Hyperbolas? Hyperbolas can be thought of as two parabolas going in opposite directions
Hyperbola For any point P that is on the hyperbola, d2 – d1 is always the same. P d2 d1 F1 F2 In this example, the origin is the center of the hyperbola. It is midway between the foci.
Hyperbolas 1. A line through the foci intersects the hyperbola at two points, called the vertices. 2. The segment connecting the vertices is called the transverse axis of the hyperbola. V V F C F 3. The centerof the hyperbola is located at the midpoint of the transverse axis. 4. As x and y get larger the branches of the hyperbola approach a pair of intersecting lines called the asymptotes of the hyperbola. These asymptotes pass through the center of the hyperbola.
The Standard Equation of a Hyperbola With Center (0, 0) and Foci on the x-axis The equation of a hyperbola with the centre (0, 0) and foci on the x-axis is: B (0, b) The length of the transverse axis is 2a. The length of the conjugate axis is 2b. The vertices are (a, 0) and (-a, 0). The foci are (c, 0) and (-c, 0). The slopes of the asymptotes are (-c, 0) (c, 0) A2 A1 (a, 0) (-a, 0) F1 F2 B (0, -b) The equations of the asymptotes
The Standard Equation of a Hyperbola with Center (0, 0) and Foci on the y-axis The equation of a hyperbola with the centre (0, 0) and foci on the y-axis is: F1(0, c) The length of the transverse axis is 2a. The length of the conjugate axis is 2b. The vertices are (0, a) and ( 0, -a). The foci are (0, c) and (0, -c). The slopes of the asymptotes are A1(0, a) B2(b, 0) B1(-b, 0) A2(0, -a) F2(0, -c) The equations of the asymptotes
Analyzing an Hyperbola State the coordinates of the vertices, the coordinates of the foci, the lengths of the transverse and conjugate axes, and the equations of the asymptotes of the hyperbola defined by each equation. The equations of the asymptotes are For this equation, a = 2 and b = 4. The length of the transverse axis is 2a = 4. The length of the conjugate axis is 2b = 8. The vertices are (2, 0) and (-2, 0): c2 = a2 + b2 = 4 + 16 = 20 The coordinates of the foci are
Analyzing an Hyperbola For this equation, a = 5 and b = 3. The length of the transverse axis is 2a = 10. The length of the conjugate axis is 2b = 6. The vertices are (0, 5) and (0, -5): c2 = a2 + b2 = 25 + 9 = 34 The coordinates of the foci are The equations of the asymptotes are
The Standard Form of the Hyperbola with Centre (h, k) When the transverse axis is vertical, the equation in standard form is: The centre is (h, k). The transverse axis is parallel to the y-axis and has a length of 2a units. The conjugate axis is parallel to the x-axis and has a length of 2b units. The slopes of the asymptotes are (h, k) The general form of the equation is Ax2 + Cy2 + Dx + Ey + F = 0.
The Standard Form of the Hyperbola with Centre (h, k) [cont’d] When the transverse axis is horizontal, the equation in standard form is: The transverse axis is parallel to the x-axis and has a length of 2a units. The conjugate axis is parallel to the y-axis and has a length of 2b units. The slopes of the asymptotes are
Describe the hyperbola • Transverse axis is vertical • Centered at (-1,3) • Distance to vertices from center = 2 units (up & down) (-1,5) & (-1,1) • Asymptotes pass through the (-1,3) with slopes = 2/3, -2/3 • Foci units up & down from the center ,
Finding the Equation of a Hyperbola The centre is (2, 3), so h = 3 and k = 2. The transverse axis is parallel to the y-axis and has a length of 10 units, so a = 5. The conjugate axis is parallel to the x-axis and has a length of 6 units, so b = 3. The vertices are (-2, 8) and (-2, -2). The slope of one asymptote is , so a = 5 and b = 3: The coordinates of the foci are c2 = a2 + b2 = 25 + 9 = 34 Standard form
Writing the Equation in General Form 9(y - 2)2 - 25(x - 3)2 = 225 9(y2 - 4y + 4) - 25(x2 - 6x + 9) = 225 9y2 - 36y + 36 - 25x2 - 150x - 225 = 225 -25x2 +9y2 - 150x - 36y + 36 - 225 = 225 -25x2 + 9y2 - 150x - 36y - 414 = 0 The general form of the equation is -25x2 + 9y2 - 150x - 36y + 36 = 0 where A = -25, C = 9, D = -150, E = -36, F = 36.
Writing the Equation of a Hyperbola Write the equation of the hyperbola with centre at (2, -3), one vertex at (6, -3), and the coordinates of one focus at (-3, -3). The centre is (2, -3), so h = 2, k = -3. The distance from the centre to the vertex is 4 units, so a = 4. The distance from the centre to the foci is 5 units, so c = 5. Use the Pythagorean property to find b: b2 = c2 - a2 = 25 - 16 = 9 b = ±3 9(x - 2)2 - 16(y + 3)2 = 1 9(x2 - 4x + 4) - 16(y2 + 6y + 9) = 144 9x2 - 36x + 36 - 16y2 - 96y - 144 = 144 9x2 - 16y2 - 36x - 96y + 36 - 144 = 144 9x2 - 16y2 - 36x - 96y - 216 = 0 General form Standard form
Analyzing an Hyperbola State the coordinates of the vertices, the coordinates of the foci, the lengths of the transverse and conjugate axes and the equations of the asymptotes of the hyperbola defined by 4x2 - 9y2 + 32x + 18y + 91 = 0. 4x2 - 9y2 + 32x + 18y + 91 = 0 (4x2 + 32x ) + (- 9y2 + 18y) + 91 = 0 4(x2 + 8x + ____) - 9(y2 - 2y + _____) = -91 + _____ + _____ 16 1 64 -9 4(x + 4)2- 9(y - 1)2 = -36 3.5.14
General Information about HyperbolasCentered at Origin • x and y are both squared • Equation always equals(=) 1 • Equation is always minus(-) • a2 is always the first denominator • c2 = a2 + b2 • c is the distance from the center to each foci on the major axis • a is the distance from the center to each vertex on the major axis
More General Information about Hyperbolas Centered at the Origin • b is the distance from the center to each midpoint of the rectangle used to draw the asymptotes. This distance runs perpendicular to the distance (a). • Major axis has a length of 2a • Eccentricity(e): e = c/a (The closer e gets to 1, the closer it is to being circular • If x2 is first then the hyperbola is horizontal • If y2 is first then the hyperbola is vertical.
What changes if the center is NOT at the origin? • Standard Form of Equation • Equations of Asymptotes
Hyperbola © Jill Britton, September 25, 2003 The huge chimney of a nuclear power plant has the shape of a hyperboloid, as does the architecture of the James S. McDonnell Planetarium of the St. Louis Science Center.
An explosion is recorded by two microphones that are two miles apart. M1 received the sound 4 seconds before M2. assuming that sound travels at 1100 ft/sec, determine the possible locations of the explosion relative to the locations of the microphones. E(x,y) Let us begin by establishing a coordinate system with the origin midway between the microphones d2 d1 Since the sound reached M2 4 seconds after it reached M1, the difference in the distances from the explosion to the two microphones must be M2 M1 (-5280, 0) (5280, 0) 1100(4) = 4400 ft wherever E is This fits the definition of an hyperbola with foci at M1 and M2 x2 y2 – = 1 Since d2 – d1 = transverse axis, a = 2200 a2 b2 c2 = a2 + b2 52802 = 22002 + b2 The explosion must be on the hyperbola b2 = 23,038,400 x2 y2 – = 1 4,840,000 23,038,400