Hyperbola

Definition:

A hyperbola is the set of points in the plane such that for each point, the absolute value of the difference of its distances (focal radii) from two fixed points (foci) is a given constant.  The point bisecting the line segment joining the foci is called the center of the ellipse.

To measure the ovalness of an ellipse, the eccentricity is figured.
e = c/a, where c is the distance from the center to the foci, and a is the distance from the center to the vertices on the major axis.
 

Classifying Conics from General Equations:

Ax2 + Cy2 + Dx + Ey + F = 0

1.  Circle:  A = C
2.  Parabola::  AC = 0
3.  Ellipse:  AC > 0
4.  Hyperbola:  A < C
 

Derive the equation of a hyperbola:

Let the center of the ellipse be (0,0), the foci of the ellipse be (c,0) and (-c,0) and the constant difference of the focal radii be 2a where a > 0. (x,y) is a point on the ellipse.
According to the definition,  \/((x + c)2 + (y - 0)2)  -  \/((x - c)2 + (y - 0)2) = 2a
Simplifying the radicands,    \/(x2 + 2cx + c2 + y2)  -  \/(x2 - 2cx + c2 + y2 = 2a
Subtract one radicand,        \/(x2 + 2cx + c2 + y2) = 2a + \/(x2 - 2cx + c2 + y2)
Square both sides,              x2 + 2cx + c2 + y2 = 4a2 + 4a \/(x2 - 2cx + c2 + y2) + x2 - 2cx + c2 + y2
Combine like terms,            4cx = 4a2 + 4a \/(x2 - 2cx + c2 + y2)
Subtract 4a2,                     4cx - 4a2 =  4a \/(x2 - 2cx + c2 + y2)
Divide by - 4a,                    cx - a = \/(x2 - 2cx + c2 + y2)
                                            a
Get common denominator,  cx - a2  =  \/(x2 - 2cx + c2 + y2)
                                              a
Square both sides,              a4 - 2a2cx + c2x2   =   x2 - 2cx + c2 + y2
                                                                         a2
Multiply by a2,                  a4 - 2a2cx + c2x2   =   a2x2 - 2a2cx + a2c2 + a2y2
Combine like terms,           c2x2 - a2x2 - a2y2 =  a2c2 - a4
Factor,                              (c2 - a2)x2 - a2y2 = a2(c2 - a2)
Divide by a2(c2 - a2),              x2 -      y2      =   1
                                               a2    c2 - a2
Since a > c, reverse the denominator,   x2   -       y2      =   1
                                                           a2         c2 - a2
Let b2 = c2 - a2, and                            x2 -   y2   =   1
                                                            a2     b2

Things to remember about  x2 -   y2   =   1
                                        a2      b2
1.  x-intercepts are (a,0) and (-a,0) (transverse axis)
2.  y-intercepts of the box used for graphing are (0,b) and (0,-b)  (conjugate axis)
3.  a2 + b2 = c2
4.  Foci are (c,0) and (-c,0)
5.  Difference of focal radii is 2a
6.  The major axis is on the x-axis (horizontal)
7.  Center is (0,0)
8.  Asymptotes:  y = +(b/a)x

Things to remember about y2 -   x2   =   1
                                        a2      b2
1.  x-intercepts are (a,0) and (-a,0)
2.  y-intercepts are (0,b) and (0,-b)
3.   a2 + b2 = c2
4.  Foci are (0,c) and (0,-c)
5.  Difference of focal radii is 2a
6.  The major axis is on the y-axis (vertical)
7.  Center is (0,0)

       (x - h)2 -   (y- k)2   =   1    follows the same patterns but the center is now (h,k), the intercepts are (h + a, k),
          a2             b2                                  and foci are (h + c, k). Asymptotes are y - k = +(b/a)(x - h)

       (y - k)2 -   (x- h)2   =   1    follows the same patterns but the center is now (h,k), the intercepts are (h, k + a),
         a2             b2                        and foci are (h, k + c).  Asymptotes are y - k = +(b/a)(x - h)
 

Examples:
1.  Find the standard equation for the hyperbola having foci at (1,2) and (5,2) and a major axis of length 4.

     The center is the midpoint of the foci: (3,2)
     The distance from the center to the foci is 2, so c = 2.
     The major axis is 4, meaning the sum of the radii is 4, so a = 4.

2.  Graph  x2 - 4y2 + 4x - 16y - 28 = 0

     Complete the square so the center and intercepts can be seen.
      x2 + 4x - 4y2  - 16y  = 28
      x2 + 4x - 4(y2  + 4y)  = 28
      (x2 + 4x + 4) - 4(y2 + 4y + 4)  = 28 + 4 - 16  (Remember to multiply the second 4 by 2 since it is factored)
      (x + 4)2 - 4(y + 2)2 = 16  (Divide by 16)
      (x + 2)2 - (y + 2)2 =  1
         16             4

      This is a horizontal ellipse with center (-2,-2) and transverse vertices (2,-2), (-4,-2), and
      conjugate vertices (-2,0), (-2, -4).
      Plot the vertices and form a box with the vertices as midpoints of the four sides.  Draw the diagonals of the box using
      dotted lines.  These are the asymptotes of the hyperbola.  This is a horizontal hyperbola with vertices touching the sides of
      the box.  Draw the hyperbola so that the curve touches the midpoint of the sides of the box and approaches the
      asymptotes.

3.  Find the center, vertices, foci, asymptotes, and eccentricity of the hyperbola in #2.
     The center is  (-2,-2), the vertices are (-2,2), (-6,2).  a = 4.  b = 2
     Since a2 + b2 = c2 ,                                   eccentricity = c/a = 2\/5 / 5
              16 + 4 = c2                                      asymptotes:  y + 2 = +(1/2)(x - 4)
                     20 = c2
                   2\/5 = c
       So the foci are (-2 + 2\/5, 2)

4.  Classify each graph:
     (a)  2x2 - 3x + 4y - 2 = 0                      Answer:  parabola        AC = 0
     (b)  3x2 - 5y2 + 7x - 3y + 1 = 0           Answer:  hyperbola       AC < 0
     (c)  5x2 + 3y2 - 5x + 10y = 0               Answer:  ellipse             AC > 0
     (d)  2x2 + 2y2 - 5x + 7y - 4 = 0           Answer:  circle               A = C

Problems (there are no problems here yet)

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