Mini-Project for
Calculus and Mathematica
Workshop

Conics

by Marian Robinson

This lesson treats conics as graphs of
a general second degree equation in two variables and relates each of them to their respective standard equation form through the concept of locus.  Each conic illustrated becomes an integral part of the design above.  Other treatment  of conics mentioned will be developed at a later time.

1.  The General Definition of a Conic

The study of Conics  (the circle, ellipse, parabola, and hyperbola) in mathematics has always been a very fascinating  topic and often manifests itself in many forms.  Because of this, at this writing, only a number of special cases shall be considered. 
      Algebraically,  the general equation of any conic can be written in the following form:
                     (Ax^2)+ Bxy + C(y^2) + Dx + Ey + F = 0,
                       where     (A,B,C)  ≠  (0,0,0).
     The  choice of a specific conic will result from particular assignment values to the coefficients. 
      Geometrically,  the graph of this equation is called a conic section because it can be obtained by cutting a right circular cone with a plane.   These intersections are classified as circle, hyperbola,  parabola or ellipse,  each determined by the orientation of the plane to the axis or to the slant height of the cone.

2.  Case I:  The Circle

2a.  Geometric Interpretation

If a right circular cone is cut by a plane such that the plane 
is perpendicular to its axis,  then the section formed is called a circle. 
If the intersection  should occur at the vertex, then the circle is 
called a point circle, having a radius of zero.
     At a later time,  graphics illustrating this phenomena will follow.

2b:  Algebraic  Interpretation

Algebraically, the general equation of any conic can be written 
 in the following form:
                     A(x^2) + Bxy + C(y^2) + Dx + Ey + F = 0,
                       where     (A,B,C)  ≠  (0,0,0).
     The  choice of a specific conic will result from particular 
assignment values to the coefficients.

2b.1  In Factored Form

The definition of a circle is the set of all points,  P(x,y),  in a plane 
which are at a fixed distance r,  (r= radius) from a fixed point, 
C(h,k),  called the center.  In other words,  r = PC.  Double click on the
box at the right to see such a circle, with C(h,k) = (1,-1) and radius = 2.

circle = Graphics[Circle[{h, k}, r]]{h = 1, k = -1, r = 2}center = 
      Graphics[{RGBColor[1, 0, 0], PointSize[ .02], Point[{h, k}]}]Show[circle, center, AxesOrigin -> {0, 0}, 
      PlotRange -> {{-2, 4}, {-4, 2}}, 
      AspectRatio -> Automatic, 
      PlotLabel -> {"Circle with center (1,-1),radius = 2"}];

Since PC = r by definition,  and using the distance formula 
for PC , we have  sqrt [(((x  -  h) )^2)+ (((y  -  k))^2)] = r  or  by squaring, 
(((x  -  h))^2) + (((y  -  k))^2) = (r^2).   Recall,  the center is (h,k) and the 
radius is r.  Let's choose a specific center (h,k) = (2,1) and 
a radius of r = 1.  This information produces the equation:
                             (((x  -  2))^2) + (((y  -  1))^2) = 1.
                    Sketch the graph of this equation.

Clear[x, y, h, k, r]{h = 2, k = 1, r = 1}\(circle = Graphics[Circle[{h, k}, r]]; 

    center = Graphics[PointSize[ .02], Point[{h, k}]]; 

    rteye = Show[circle, center, AxesOrigin -> {0, 0}, 
		
        Axes -> True, PlotRange -> {{\(-8\), 8}, {\(-8\), 8}}, 
        AspectRatio -> Automatic]; \)

Another equation could be:  (x^2) + (y^2) = 36.  That means, the center
is (0,0), and the radius = 6.  Let's see the size and position of this 
graph in the coordinate plane.

Clear[h, k, r]{h = 0, k = 0, r = 6}\(circle = Graphics[Circle[{h, k}, r]]; 

    center = Graphics[PointSize[ .02], Point[{h, k}]]; 

    face = Show[circle, Axes -> {0, 0}, 
        PlotRange -> {{\(-8\), 8}, {\(-8\), 8}}, 
        AspectRatio -> Automatic]; \)

What happens if a radius should equal a zero value? Let's try it.
 r = 0,  and choose any center, say (h,k) = (0,-1).

Clear[h, k, r]{h = 0, k = -1, r = 0}\(circle = Graphics[Circle[{h, k}, r]]; 

    center = Graphics[PointSize[ .02], Point[{h, k}]]; 

    nose = Show[circle, center, Axes -> {0, 0}, 
        PlotRange -> {{\(-8\), 8}, {\(-8\), 8}}, 
        AspectRatio -> Automatic]; \)

Did you expect that point which appeared on the graph?    This 
is commonly called a point circle,  which in this case was located at (0,-1).

2b.2  In Expanded Form

In factored form, its representation is
        (((x  -  h))^2) + (((y  -  k))^2) = (r^2),     with (h,k) = center, r = radius.
In expanded form, the circle is represented by the general equation 
        A(x^2) + Bxy + C(y^2) + Dx + Ey + F = 0   where B = 0, and A = C. 
This form is obtained using several algebraic operations.  Squaring 
and simplifying, we arrive at  
                        (x^2) + y\(TraditionalForm\`\(\ \^2\)\)- 2h x - 2k y + h\(TraditionalForm\`\(\ \^2\)\) + k\(TraditionalForm\`\(\ \^2\)\)- r\(TraditionalForm\`\(\ \^2\)\)   = 0     
By letting 
                                      D = -2h
                                      E =  - 2k
                              and  F  = (h\(TraditionalForm\`\(\ \^2\)\)+ k\(TraditionalForm\`\(\ \^2\)\) - r\(TraditionalForm\`\(\ \^2\)\))
                               
the more familiar form of the equation  appears:
                               x\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\)+ Dx + Ey + F = 0.        
It should be noted that if  
     (D\(TraditionalForm\`\(\ \^2\)\))/4 + (E\(TraditionalForm\`\(\ \^2\)\))/4 - F > 0,  then we have a circle,  or
          (D\(TraditionalForm\`\(\ \^2\)\))/4 + (E\(TraditionalForm\`\(\ \^2\)\))/4 - F = 0,  then we have a point circle,  or     
              (D\(TraditionalForm\`\(\ \^2\)\))/4 + (E\(TraditionalForm\`\(\ \^2\)\))/4 - F < 0,   then we have an imaginary circle.

Suppose one is given an equation in expanded form such as: 
                               x\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\)+ 4x - 2y + 4 = 0.
 From these coefficients,  determine  the coordinates of the center 
 and the radius.  Since D = 4, E = -2 and F = 4, the center and radius
 are (-2,1) and 1, respectively.    (An alternative method would be 
 to complete the square in both x and y.  This will also yield the 
 standard form of the equation of the circle so that the center and 
 radius can easily be determined. )   Either way,  the standard form of 
 the equation is (x + 2)\(TraditionalForm\`\(\ \^2\)\) + (y - 1)\(TraditionalForm\`\(\ \^2\)\)= 1, and the its graph is as  follows. 
 At a later time, a more intense discussion will follow.

Clear[x, y, h, k, r]{h = -2, k = 1, r = 1}circle = Graphics[Circle[{h, k}, r]]; center = Graphics[PointSize[ .02], Point[{h, k}]]; lefteye = 
      Show[circle, center, AxesOrigin -> {0, 0}, Axes -> True, 
        PlotRange -> {{-8, 8}, {-8, 8}}, 
        AspectRatio -> Automatic];

2b.3  Using Eccentricity

This section will be introduced at a later date.

2b.4  In Rotated Form

Using,  x = x' cos t - y' sin t,  and
                y = x' sin t + y' cos t,  
where t is the angle of rotation,  then the general form 
           Ax\(TraditionalForm\`\(\ \^2\)\)+ Bxy + C\(TraditionalForm\`\(\ \^2\)\) + Dx + Ey + F = 0,  for (A,B,C) ≠ (0,0,0),  
becomes
           A'(x')\(TraditionalForm\`\(\ \^2\)\) + B'(x'y') + C'(y'(() )^2) + D'(x') + E'(y') + F' = 0 
where A' = A cos\(TraditionalForm\`\(\ \^2\)\)t + B cos t sin t + C sin\(TraditionalForm\`\(\ \^2\)\)t   + B cos ø sin ø + C sin\(TraditionalForm\`\(\ \^2\)\)ø,
           B' = 2 (C - A) sin t cos t + B [cos\(TraditionalForm\`\(\ \^2\)\)- sin\(TraditionalForm\`\(\ \^2\)\)],
           C' = A sin\(TraditionalForm\`\(\ \^2\)\)t -  B cos t sin t + C cos\(TraditionalForm\`\(\ \^2\)\)t
           D' = D cos t + E sin t
           E' = - D sin t + E cos t
           F' = F.
           Substitute an angle of rotation, and the new  general form 
 can be determined.  Further discussion will follow at a later date.

2b.5  In Polar Form

This form will be developed later.

2b.6  In Parametric Form

This form will be developed later.

2b.7  Pictoral Summary for the Circles

3.  Case II: The Ellipse

3a.  The Geometric Interpretation

If a plane is passed through a right circular cone in such a manner that it is oblique to its axis and not parallel to the slant height, then the section determined is called an ellipse.
     At a later date,  suitable geometric graphics illustrating this concept will follow.

3b.  The Algebraic Interpretation

Definition:  The set of all points, P(x,y), in a plane such that the sum of the distances from P(x,y) to two fixed points F¡(-c,0) and F(c,o), (called foci)  of the plane is a constant.  For the time being, let's call that constant 2a. In other words, 
                                      PF¡ + PF = 2a.

3b.1  In Factored Form

With point P(x,y),  and foci  F¡ (-c,0), and F(c,0)  in conjunction with the distance formula,  calculate the following:
                                 PF' + PF = 2a.
                                   
Hence,         sqrt   [(x + c) \(TraditionalForm\`\(\ \^2\)\)+ y \(TraditionalForm\`\(\ \^2\)\)] +  sqrt [(x - c)\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\)] = 2a.
If, now,  - sqrt[(x + c)  \(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\)] is added to both members of this equation, one obtains
        (x + c)\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\) = 4a\(TraditionalForm\`\(\ \^2\)\) - 4a sqrt[(x - c)\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\)] + (x - c)\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\).
Expanding binomials and simplifying will yield
                            sqrt[(x - c)\(TraditionalForm\`\(\ \^2\)\)+ y\(TraditionalForm\`\(\ \^2\)\)] a = a\(TraditionalForm\`\(\ \^2\)\)- cx.
If both members of this equation are squared and equated,                                
                 a\(TraditionalForm\`\(\ \^2\)\)(x\(TraditionalForm\`\(\ \^2\)\) - 2cx + c\(TraditionalForm\`\(\ \^2\)\) + y\(TraditionalForm\`\(\ \^2\)\)) = (a^4) - 2a\(TraditionalForm\`\(\ \^2\)\)cx + c\(TraditionalForm\`\(\ \^2\)\)x\(TraditionalForm\`\(\ \^2\)\).
Simplifying and factoring obtains
                        (a\(TraditionalForm\`\(\ \^2\)\) - c\(TraditionalForm\`\(\ \^2\)\))x\(TraditionalForm\`\(\ \^2\)\) + a\(TraditionalForm\`\(\ \^2\)\)y\(TraditionalForm\`\(\ \^2\)\) = a\(TraditionalForm\`\(\ \^2\)\)(a\(TraditionalForm\`\(\ \^2\)\) - c\(TraditionalForm\`\(\ \^2\)\)).
Set b\(TraditionalForm\`\(\ \^2\)\) = a\(TraditionalForm\`\(\ \^2\)\) - c\(TraditionalForm\`\(\ \^2\)\), b > 0, then this last equation reduces to
                                ((((x )^2))/((a )^2))+((((y )^2))/((b )^2))= 1.
    (Note:  if the foci F'(0,c) and F(0,-c) are chosen to lie on the y-axis, then using similar calculations reduces the equation to
                               ((((x )^2))/((a )^2))+((((y )^2))/((b )^2))= 1.)
Let's investigate the notion of translation.

If an appropriate translation were used, say
                    x' - h = x    and    y' - k = y
then the equations become, respectfully:  
(((((x'  -  h) )^2))/((a )^2))+(((((y'  -  k) )^2))/((b )^2))= 1   or  (((((x'  -  h) )^2))/((b )^2))+(((((y'  -  k) )^2))/((a )^2))= 1.                   
Note:         (h,k) represents the center of the ellipse where
                  a = length of semi-major axis 
           and b = length of semi-minor axis. This is referred to as the standard form of an ellipse. Now, it is time for a few concrete examples.  Graph the following equation in this standard form. 
           (((((x  -  7) )^2))/1)+((((y )^2))/9)= 1.

Clear[x, y, h, k, a, b]f[h_, k_, a_, b_] := {(Sqrt[(b^2) - (((b^2) (((x - h))^2))/(a^2))]) + k, 
        -(Sqrt[(b^2) - (((b^2) (((x - h))^2))/(a^2))]) + k}f[-7, 0, 1, 3]leftear = 
      Plot[Release[f[-7, 0, 1, 3]], {x, -8, -6}, 
        Axes -> True, AxesOrigin -> {0, 0}, 
        AspectRatio -> Automatic];

Graph the following equation in this standard form.   
           (((((x  -  2) )^2))/4)+(((((y + 7) )^2))/1)= 1.

Clear[x, y, h, k, a, b]f[h_, k_, a_, b_] := {(Sqrt[(a^2) - (((a^2) (((x - h))^2))/(b^2))]) + k, 
        -(Sqrt[(a^2) - (((a^2) (((x - h))^2))/(b^2))]) + k}f[2, -7, 1, 2]rtbowtie = 
      Plot[Release[f[2, -7, 1, 2]], {x, 0, 4}, Axes -> True, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

3b.2  In Expanded Form

If the standard form of the equation of the ellipse is expanded and 
terms are regrouped, then the equation of the ellipse takes on the form
                       Ax\(TraditionalForm\`\(\ \^2\)\) + Cy\(TraditionalForm\`\(\ \^2\)\) + Dx + Ey + F = 0, 
                       where A and C are not equal,   and AC > 0.
(Obviously, completing the square in x and y reverts the general form
back to the standard form.)
A few examples are in order here, too.  Let's try the following:
     9x\(TraditionalForm\`\(\ \^2\)\) + y\(TraditionalForm\`\(\ \^2\)\) - 108x + 315 = 0.  By completing the square, we arrive at
                            (x - 6)\(TraditionalForm\`\(\ \^2\)\) +  y\(TraditionalForm\`\(\ \^2\)\) = 1.
                                1            9                                                                                               
This says that the center is at (6,0) and the graph will be taller
 than it is wide.  The graph is as follows.

Clear[x, y, h, k, a, b]f[h_, k_, a_, b_] := {(Sqrt[(b^2) - (((b^2) (((x - h))^2))/(a^2))]) + k, 
        -(Sqrt[(b^2) - (((b^2) (((x - h))^2))/(a^2))]) + k}rtear = 
      Plot[Release[f[7, 0, 1, 3]], {x, 6, 8}, AxesOrigin -> {0, 0}, 
        AspectRatio -> Automatic];

Another example should reinforce this concept. 
This time, use the equation:
                          x\(TraditionalForm\`\(\ \^2\)\) + 4y\(TraditionalForm\`\(\ \^2\)\) + 4x + 56y + 196 = 0.
 After completing the square in x and y, the standard form becomes:
         (((((x  + 2) )^2))/4)+(((((y + 7) )^2))/1)= 1.
  Let's let the computer sketch its graph quickly.

Clear[x, y, h, k, a, b, leftbowtie]f[h_, k_, a_, b_] := {(Sqrt[(a^2) - (((a^2) (((x - h))^2))/(b^2))]) + k, 
        -(Sqrt[(a^2) - (((a^2) (((x - h))^2))/(b^2))]) + k}f[-2, -7, 1, 2]leftbowtie = 
      Plot[Release[f[-2, -7, 1, 2]], {x, -4, 0}, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

3b.3  Using Eccentricity

This section will be developed at a later date.

3b.4  In Rotated Form

This section will be developed at a later date.

3b.5  In Polar Form

This section will be developed at a later date.

3b.6  In Parametric Form

This section will be developed at a later date.

3b.7  Graphical Summary of Ellipses and Circles

Show[rtear, leftear, rtbowtie, leftbowtie, rteye, lefteye, face, 
      nose];

4.  Case III:  Parabola

4a.  The Geometric Interpretation

If a plane is passed through a right circular cone parallel to its
slant height, then the section created is called a parabola.  The 
degenerate case is a straight line.  The visual interpretation of this
will be developed at a later date.

4b.  The Algebraic Interpretation

Definition:  A set of all points P(x,y) in a plane that are equidistant from a fixed point F (called a focus) and a fixed line (called a directrix)  of the plane.  PF = PM, where PM is the distance from the point P to the line.

4b.1  In Factored Form

Let's examine the case where the directrix is vertical.
Using the distance formula on PF = PM,  for P(x,y) and  F(a,0), and M(-a,y), then we arrived at:
                   sqrt(x + a)\(TraditionalForm\`\(\ \^2\)\) = sqrt ((x - a)\(TraditionalForm\`\(\ \^2\)\) + y\(TraditionalForm\`\(\ \^2\)\))
Squaring both sides and collecting like terms yields
                                     4ax = y\(TraditionalForm\`\(\ \^2\)\)
Note:  a > 0    causes the parabola to open to the right.
          a < 0    causes the parabola to open to the left.
      While examining the case where the directrix is horizontal, this similar equation will follow for P(x,y), F(0,a), and M(x,-a):   
                                        4ay = x\(TraditionalForm\`\(\ \^2\)\)
 Note:  a > 0    causes the parabola to open upward
           a < 0    causes the parabola to open downward.
      The traditional translation formulas
                                      x = x' - h 
                                      y = y' - k,
  will yield the standard form of the equation.
           4a(x' - h) = (y' - k)\(TraditionalForm\`\(\ \^2\)\)    or    4a(y' - k) = (x' - h)\(TraditionalForm\`\(\ \^2\)\)
           A few sample equations, in standard form, will follow for a > 0 or a < 0.   First, try a > 0.  So let a = .01  for   
                       (y + 3) = 4a (x )\(TraditionalForm\`\(\ \^2\)\), where a > 0.

Clear[x, y, h, k, a]f[h_, k_, a_] := 4 a (((x - h))^2) + kf[0, -3,  .01]upperlip = 
      Plot[Release[f[0, -3,  .01]], {x, -2.5, 2.5}, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

Another example is in order.  Try the following with a < 0.
                             y - 4 = 4 a (x - 2)\(TraditionalForm\`\(\ \^2\)\), where a = -.5

Clear[x, y, h, k, a]f[h_, k_, a_] := 4 a (((x - h))^2) + kf[2, 5, - .5]rtbrow = 
      Plot[Release[f[2, 5, - .5]], {x, 1, 3}, AxesOrigin -> {0, 0}, 
        AspectRatio -> Automatic];

4b.2  In Expanded Form

Starting with the standard form and expanding, the general form becomes
             Ax\(TraditionalForm\`\(\ \^2\)\)+ Cy\(TraditionalForm\`\(\ \^2\)\)+ Dx + Ey + F = 0, where A or C = 0,
                                 but not both. 
In the usual manner, completing the square in the quadratic variable, we arrive at a traditional standard form.
Let's try an example, given in general form.  
                                  .2 x\(TraditionalForm\`\(\ \^2\)\) - y - 3 = 0.
A transformation to the standard form establishes  a > 0 to be equal to .05 and 
                                y + 4 = 4 a (x - 0)\(TraditionalForm\`\(\ \^2\)\).

Clear[x, y, h, k, a]f[h_, k_, a_] := 4 a (((x - h))^2) + kf[0, -4,  .05]lowerlip = 
      Plot[Release[f[0, -4,  .05]], {x, -2.5, 2.5}, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

A final example is below. 
                           -1.2 x\(TraditionalForm\`\(\ \^2\)\) - 4.8x - y - 8.8 = 0
 This time, determine that  a < 0,  such that a = -.3, from the standard form 
                                   y + 4 = 4 a (x + 2)\(TraditionalForm\`\(\ \^2\)\)

Clear[x, y, h, k, a]f[h_, k_, a_] := 4 a (((x - h))^2) + kf[-2, 4, - .3]leftbrow = 
      Plot[Release[f[-2, 4, - .3]], {x, -3, -1}, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

To get a parabola to open to the right (a > 0), use the form
                                    ( x - h) = 4a (h - k)\(TraditionalForm\`\(\ \^2\)\)
              Let's try one.     x - 1 = 4a (y - 7)\(TraditionalForm\`\(\ \^2\)\)with a = .27

Clear[x, y, h, k, a]f[h_, k_, a_] := {2 (Sqrt[(x - h) a]) + k, 
        -2 (Sqrt[(x - h) a]) + k}f[-1, 9,  .27]leftcurl = 
      Plot[Release[f[-1, 9,  .27]], {x, -1, 0}, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

To get a parabola to open to the left (a < 0), use the form
                                    ( x - h) = 4a (h - k)\(TraditionalForm\`\(\ \^2\)\)
              Let's try one.     x - 1 = 4a (y - 7)\(TraditionalForm\`\(\ \^2\)\) with a = -.26

Clear[x, y, h, k, a]f[h_, k_, a_] := {2 (Sqrt[(x - h) a]) + k, 
        -2 (Sqrt[(x - h) a]) + k}f[1, 7, - .26]rtcurl = 
      Plot[Release[f[1, 7, - .26]], {x, 0, 1}, AxesOrigin -> {0, 0}, 
        AspectRatio -> Automatic];

4b.3  Using Eccentricity

This section will be developed at a later date.

4b.4  In Rotated Form

This section will be developed at a later date.

4b.5  In Polar Form

This section will be developed at a later date.

4b.6  In Parametric Form

This part will be developed at a future date.

5.  Case IV:  The Hyperbola

5a.  The Geometric Interpretation

For a discussion of the hyperbola, it is necessary to use both nappes
of a right circular cone.  The section formed by passing a plane
parallel to the axis, and cutting both nappes is called a hyperbola.
The degenerate case is a pair of intersecting lines.  A further discussion
will be forthcoming at a later date.

5b.  The Algebraic Interpretation

The hyperbola is similar to the ellipse in that instead
of the sum, one has the difference of the distances 
PF¡ and PF = 2a.   That is, hyperbola is the set of points in a plane
such that the difference of the distances from two fixed points (foci) 
in the plane is constant, in particular = 2a, again.

5b.1  In Factored Form

The standard form of the equation when developed similarly is:
           
         (x - h)\(TraditionalForm\`\(\ \^2\)\)-  (y - k)\(TraditionalForm\`\(\ \^2\)\) = 1              (y - k)\(TraditionalForm\`\(\ \^2\)\)- (x - h)\(TraditionalForm\`\(\ \^2\)\) = 1
             a\(TraditionalForm\`\(\ \^2\)\)           b\(TraditionalForm\`\(\ \^2\)\)             or            a\(\ \^2\)           b \(TraditionalForm\`\^2\)         
         with vertex on a line                with vertex on a line
         parallel to the x-axis                 parallel to the y-axis
         
  For this lesson, one example will be given.  The same equation will
  be:         x\(\ \^2\)- (y + 2)\(\ \^2\) = 1.
                 9        4

Clear[x, y, h, k, a, b]f[h_, k_, a_, b_] := {
        (Sqrt[-(b^2) + (((b^2) (((x - h))^2))/(a^2))]) + k, 
        -(Sqrt[-(b^2) + (((b^2) (((x - h))^2))/(a^2))]) + k}f[0, -3, 2.5, 1]cheeks = 
      Plot[Release[f[0, -3, 2.5, 1]], {x, -3.5, 3.5}, 
        AxesOrigin -> {0, 0}, AspectRatio -> Automatic];

5b.2  In Expanded Form

If the standard form of the equation of the hyperbola is expanded 
and terms are regrouped, then the equation of the hyperbola takes on 
the form
                       Ax\(\ \^2\) + Cy\(\ \^2\)+ Dx + Ey + F = 0, 
                       where A and C are not equal, 
                       and AC<0.
(Obviously, completing the square in x and y reverts the general form
back to the standard form.)
     At the time of this writing, no examples will be provided for this
 form.

5b.3  Using Eccentricity

This section will be developed at a future date.

5b.4  In Rotated Form

This section will be developed at a future date.

5b.5  In Polar Form

This section will be developed at a future date.

5b.6  In Parametric Form

This section will be enhanced at a future date.

6.  Graphical Summary of the 4 Conic Sections

Show[rtear, leftear, rtbowtie, leftbowtie, rteye, lefteye, face, 
      nose, cheeks, rtbrow, leftbrow, upperlip, lowerlip, rtcurl, leftcurl, 
      Axes -> False];