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THE HYPERBOLA

THE HYPERBOLA. A hyperbola is the collection of all points in the plane the difference of whose distances from two fixed points, called the foci, is a constant. This is the same definition as an ellipse except we have the difference is always constant instead of the sum.

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THE HYPERBOLA

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  1. THE HYPERBOLA

  2. A hyperbola is the collection of all points in the plane the difference of whose distances from two fixed points, called the foci, is a constant. This is the same definition as an ellipse except we have the difference is always constant instead of the sum. The hyperbola has two symmetric parts called branches. conjugate axis Each branch has a vertex and a focus (F). vertices The axis that contains the vertices is called the transverse axis. Transverse axis focus focus An axis at right angles to the transverse axis is called the conjugate axis.

  3. PARTS OF A HYPERBOLA The distance between the vertices is 2a Midpoint of the line joining the vertices is the center (C) of the hyperbola conjugate axis c Construct a rectangle with its center at C and length 2a vertex 2b C vertex focus focus 2a The width of this rectangle is 2b the conjugate axis length. transverse axis The diagonals of this rectangle are asymptotes of the segments of the hyperbola. c is the distance from the center to the vertex. c2 = a2 + b2 For a hyperbola:

  4. Standard equation of a hyperbola with its center at the origin and horizontal transverse axis For a hyperbola with its center at the origin and has the transverse axis horizontal, the standard equation is: c vertex vertex 2b C (-c, 0) (c, 0) focus (-a, 0) (a, 0) focus 2a c2 = a2 + b2 The equations of its asymptotes are: Focus: (c, 0) (-c, 0) (c, 0) vertices: (a, 0) (-a, 0) (a, 0)

  5. Standard equation of a hyperbola with its center at the origin and vertical transverse axis For a hyperbola with its center at the origin and has the transverse axis horizontal, the standard equation is: (0, c) (0, a) c2 = a2 + b2 (0, -a) The equations of its asymptotes are: (0, -c) Focus: (0, c) (0, -c) (0, c) vertices: (0, a) (0, -a) (0, a)

  6. Find the coordinates of foci, the lengths of transverse and conjugate axes and sketch the graph of This is a2 This is b2 a = 3 b = 2 Transverse axis horizontal c2 = a2 + b2 = 9 + 4 = 13 c = 13 Focus: (c, 0) (-13, 0) (13, 0) vertices: (  a, 0) (-3, 0) (3, 0) Construct the rectangle with 2a = 6 units as its length and 2b = 4 units as its width with center at the origin. 2b = 4 (-13, 0) (3, 0) (-3, 0) (13, 0) 2a = 6 The diagonals of this rectangle are the asymptotes of the hyperbola.

  7. Find the coordinates of foci, the lengths of transverse and conjugate axes and sketch the graph of This is a2 This is b2 a = 5 b = 3 Transverse axis vertical c2 = a2 + b2 = 25 + 9 = 34 c = 34 Focus: (0, c) (0, -34) (0, 34) (0, 34) vertices: (0,  a) (0, 5) (0, -5) (0, 5) Construct the rectangle with 2a = 10 units as its length and 2b = 6 units as its width with center at the origin. The diagonals of this rectangle are the asymptotes of the hyperbola. (0, -5) (0, -34)

  8. Find the coordinates of foci, the lengths of transverse and conjugate axes and sketch the graph of 4y2 – 25x2 = 100 Divide by 100 to Make the right side 1 Transverse axis vertical a = 5 b = 2 (0, 29) c = 29 c2 = a2 + b2 = 25 + 4 = 29 (0, 5) (0, 29) Focus: (0, c) (0, -29) (0, 5) (0, -5) vertices: (0,  a) Construct the rectangle with 2a = 10 units as its length and 2b = 4 units as its width with center at the origin. (0, -5) (0, -29) The diagonals of this rectangle are the asymptotes of the hyperbola.

  9. Write the equation of a hyperbola that has its transverse axis on the x-axis, length of transverse axis = 8 and the length of conjugate axis = 6 Transverse axis on the x-axis makes it horizontal 2a = 8 a = 4 2b = 6 b = 3 The equation is:

  10. Write the equation of a hyperbola that has its transverse axis on the y-axis, length of transverse axis = 16 and the distance of the foci from the center is 10 Transverse axis on the y-axis makes it vertical c = 10 2a = 16 a = 8 c2 = a2 + b2 b2 = c2 – a2 = 100 – 64 = 36 b = 6 The equation is:

  11. Standard equation of a hyperbola with its center (h, k) and horizontal transverse axis A hyperbola with center at C(h, k)is obtained by translating the hyperbola with center at the origin hunits horizontally and kunits vertically. (h, k) (h – c , k) (h + c, k) (h – a, k) (h + a, k) The equation for such a hyperbola is: Focus: (hc, k) (h + c, k) (h – c , k) vertices: (h a, k) (h + a, k) (h – a, k)

  12. Standard equation of a hyperbola with its center (h, k) and vertical transverse axis (h, k + c) A hyperbola with center at C(h, k)is obtained by translating the hyperbola with center at the origin h units horizontally and k units vertically. (h, k + a) (h, k) (h, k – a) The equation for such a hyperbola is: ( h, k – c) Focus: (h, k c) (h, k + c) ( h, k – c) vertices: (h, k a) (h, k + a) (h, k – a)

  13. Find the center, foci, vertices and plot the graph of 4x2 – 9y2– 16x –36y +16 = 0 Complete square separately on x-terms and y-terms by grouping them. Move the constant 16 to the right – 9y2 – 36y = -16 4x2 – 16x Make the coefficients of x2 and y21by factoring out 4and 9 – 9(y2+ 4y ) – 36 4(x2 – 4x ) + 22 + 22 = -16 + 16 x2 – 4x becomes a perfect square when 22 is added to it. When 22 is added, 4 × 22 = 16 gets added to the left To compensate we add 16 to the right y2 + 4y becomes a perfect square when 22 is added to it. To compensate we subtract 9 × 22 = 36 to the right 4(x – 2)2– 9(y + 2)2 = -36 Divide each term by -36 -9 4

  14. Transverse axis vertical h = 2, k = 3 Center(2, 3) a = 2 b = 3 c2 = a2 + b2 = 4 + 9 = 13 (2, 3+13) c = 13 (2, 5) Focus: (2, 3c) (2, 3) (2, 3-13) (2, 3+13) (2, 1) vertices: (2, 3 a) (2, 5) (2, 1) Construct the rectangle with 2a = 4 units as its one side and 2b = 6 units as its other side with center at the origin. (2, 3-13) The diagonals of this rectangle are the asymptotes of the hyperbola.

  15. Find the center, foci, vertices and graph the hyperbola: 1 1 1 1 Grouped the y terms and factored out a 4, grouped the x terms and factored out a -3. The y square term is first because it is positive. The right hand side must be a 1 so divide all terms by 12 1 12 12 12 4 3 This is now in standard form and we are ready to find what we need and graph.

  16. The center is at (h, k). In this case (1, -1). this is a2 so a = 3 this is b2 so b = 2 a = square root 3 so the vertices (on the transverse axis) are square root of 3 each way from the center. Since it is the y term that is positive, we move square root of 3 each way in the y direction. so vertices are: To find foci: (1, -1) So foci are: Make the rectangle and asymptotes to help you graph

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