Foci of a conic section




In this chapter we only consider affine conic sections in an orthonormal coordinate system.

Isotropic tangent lines

Theorem: The tangent lines out of a focus of a non-degenerated conic section are isotropic.

Proof:

  1. For an ellipse
    The ellipse has equation
     
            b2  x2  + a2  y2  - a2 b2  z2 = 0
    
    The quadratic equation of the tangent lines out of focus point F(c,0,1) is
     
    (c b2  x - a2  b2  z)2  - 4 (b2  c2  - a2  b2 ) (b2  x2  + a2  y2  - a2  b2  z2 ) = 0
    
            m is slope of a tangent line
    <=>
            (1,m,0) is on a tangent line
    <=>
            c2  b4  + b4  (b2  - a2  m2) = 0
    <=>
            ...
    <=>
            m2 = -1
    
  2. For an hyperbola
    Similar proof.
  3. For a parabola The parabola has equation
     
            y2  - 2 p x z = 0
    
    The quadratic equation of the tangent lines out of focus point F(p/2,0,1) is
     
             p                  2           p     2
            (- (-2 p z) - 2 p x)  - 4 (-2p (-)) (y  - 2 p x z) = 0
             2                              2
    
            m is slope of a tangent line
    <=>
            (1,m,0) is on a tangent line
    <=>
    
            4 p2  x2  + 4 p2  m2  = 0
    <=>
            m2  = -1
    

Isotropic lines and a circle

Theorem 1

The quadratic equation of the isotropic lines through a regular point P is the equation of a circle with center P.

Proof:
The isotropic lines through P(xo,yo) have slope i and -i.
The equations are

 
        y - yo = i (x - xo)  and y - yo = -i (x - xo)
<=>
        y - yo - i (x - xo) = 0 and  y - yo + i (x - xo) = 0
The quadratic equation of these lines is
 
        (y - yo - i (x - xo))(y - yo + i (x - xo)) = 0
<=>
        (y - yo)2  + (x - xo)2  = 0
This is the equation of a circle with center P.

Theorem 2

If the quadratic equation of two lines is a circle, then these lines are the isotropic lines through a regular point.

Proof:
If the quadratic equation of two lines is a circle, then the equation is

 
        (y - yo)2  + (x - xo)2  - r2  = 0
But if this is the equation of two lines, the circle is degenerated. Thus
 
        DELTA = 0  <=>  ... <=>  r = 0

Then, the equation is
 
        (y - yo)2  + (x - xo)2  = 0
Factorizing, we find
 
        (y - yo - i (x - xo))(y - yo + i (x - xo)) = 0
This is the equation of two lines through a regular point.

Foci and isotropic tangent lines

Theorem:
First take a non-degenerated real ellipse different from a circle.
The real regular points, out of which the tangent lines are isotropic, are the foci of the ellipse.

Proof:
The ellipse has equation

 
        b2  x2  + a2  y2  - a2  b2  z2 = 0
P(xo,yo,1) is a real regular point. The tangent lines through P have as equation
 
        (x.Fx'(xo,yo,1) + y.Fy'(xo,yo,1) + z.Fz'(xo,yo,1))2

                - 4 F(x,y,z).F(xo,yo,1) = 0
<=>
        (x b2  xo + y a2  yo + z(- a2  b2 ))

        - ( b2  x2  + a2  y2  - a2  b2  z2 )( b2  xo2  + a2  yo2  - a2  b2  z2 )=0


        These tangent lines are the isotropic lines
<=>
        Previous equation is a circle
<=>
        /
        |  (b2  xo2 )  - b2 ( b2  xo2  + a2  yo2  - a2 b2  z2 )
        |
        |       = (a2  yo)2  - a2  ( b2  xo2  + a2  yo2  - a2  b2  z2 )
        |
        |  b2  xo a2 yo  = 0
        \
The second condition gives xo = 0 or yo = 0.
If xo = 0, the system has no real solutions for xo and yo. (exercise)
If yo = 0, The system gives xo = c or xo = -c. (exercise)
The theorem also holds for a non-degenerated hyperbola
The proof is left as an exercise.

The theorem also holds for a non-degenerated parabola
There is only one point for which the tangent lines are isotropic.
We know that the focus is such point.
Thus, the focus is the only point with that property

Orthogonality and isotropic lines

Theorem 3

Two orthogonal lines are harmonic conjugate lines relative to the isotropic lines through their intersection point.

Proof:
Choose the orthonormal axes on the two orthogonal lines. The two isotropic lines through their intersection point are y + ix = 0 and y - ix = 0.
The equations of the four lines are :

 
        y = 0   ;  x = 0  ;  y + ix = 0  ;  y - ix = 0
We see that the condition for harmonic conjugate lines is satisfied.

Theorem 4

If two lines, b and c, are harmonic conjugate lines relative to the isotropic lines through their intersection point, then they are orthogonal.

Proof:
Choose the orthonormal axes such that the origin is in the intersection point of b and c and that line b is on the x-axis.
We call j and k the isotropic lines through the origin.
The equations of j, k and b are

 
        y + ix = 0   ;  y - ix = 0  ;  (y + ix) + 1(y - ix) = 0
The equation of the line c, harmonic conjugate to b, then is
 
        (y + ix) - 1(y - ix) = 0    <=>  x = 0
So, b and c are orthogonal.

The polar line of the focus of a parabola is the directrix

The proof is left as an exercise.

Directrix of an ellipse

The ellipse has equation
 
        b2  x2  + a2  y2  - a2  b2  z2 = 0
If we calculate the polar line of F(c,0), we find
 
            a2
        x = --
            c
We call this line the directrix of F.
If we calculate the polar line of F'(-c,0), we find
 
              a2
        x = - --
              c
We call this line the directrix of F'.

Directrix of a hyperbola

The hyperbola has equation
 
        b2  x2  - a2  y2  - a2  b2  z2 = 0
If we calculate the polar line of F(c,0), we find
 
            a2
        x = --
            c
We call this line the directrix of F.
If we calculate the polar line of F'(-c,0), we find
 
              a2
        x = - --
              c
We call this line the directrix of F'.

Eccentricity of an ellipse

Theorem:
The ratio of the distances from a point P on a non-degenerated ellipse E, to a focus F and to the corresponding directrix d is constant.

Proof:

 
                         2
             2          a  2            2
        |P,d|  =  (xo - --)   and  |P,F|  =  (xo - c)2  + yo2
                        c

with
        b2  xo2  + a2  yo2  = a2  b2

Thus

        a2  |P,F|2  =  a2  (xo - c)2  +  a2  b2  - b2 xo2

                   = ... = (c xo - a2 )2
and from this

                                  2
          2   2       2          a  2
        (a / c ) |P,F|  =  (xo - --)
                                 c
and so,

        |P,F|    c
        ----- = ---  = constant and < 1
        |P,d|    a

Similarly for the other focus.
        |P,F'|   c
        ----- = ---  = constant and < 1
        |P,d|    a
The constant value e = c/a is called the eccentricity of the ellipse.

Eccentricity of a hyperbola

Similarly as above you'll find for a hyperbola
 
        |P,F|   |P,F'|   c
        ----- = ----- = ---  = constant and > 1
        |P,d|   |P,d|    a
The constant value e = c/a is called the eccentricity of the hyperbola.
It is easy to prove that for an orthogonal hyperbola
 
                                     ___
                eccentricity = e =  V 2

Plucker lines and Plucker Hyperbolas

Say F(x,y,z) = 0 is the equation of a non-degenerated real conic section different from a circle.
 
        P(xo,yo,1) is focus of this conic section
<=>
        The tangent lines out of P are isotropic
<=>
        (x.Fx'(xo,yo,1) + y.Fy'(xo,yo,1) + z.Fz'(xo,yo,1))

                - 4 F(x,y,z).F(xo,yo,1) = 0   are isotropic
<=>
        (xo,yo) is a solution of the system
        /
        | (Fx'(x,y,1))2  - 4 a F(x,y,1) =  (Fy'(x,y,1))2  - 4 a'F(x,y,1)
        |
        |
        |  Fx'(x,y,1).Fy'(x,y,1) - 4 b" F(x,y,1) = 0
        \
<=>
        (xo,yo) is a solution of the system
        /
        |  (Fx'(x,y,1))2  -  (Fy'(x,y,1))2  = 4(a - a') F(x,y,1)
        |
        |
        |  Fx'(x,y,1).Fy'(x,y,1) = 4 b" F(x,y,1)
        \
  1. The conic section is a parabola.
    With a lot of algebraic work it can be shown that previous system is a system of linear equations. The equations are equations of lines. These two lines are called the Plucker lines of the parabola. The focus of the parabola is the intersection point of the two Plucker lines.
  2. The conic section is a ellipse or hyperbola.
    With a lot of algebraic work it can be shown that previous system is a system of hyperbolas. these hyperbolas are called the Plucker hyperbolas. To calculate the foci, we calculate the equations of the axes. Then we calculate the intersection points of the axes with a Plucker hyperbola.

Example

 
  F(x,y,1) =  6 x2  - 4 x y + 9 y2  - 4 x - 32 y - 6 = 0

  Fx'(x,y,1) = 12 x - 4 y - 4

  Fy'(x,y,1) = - 4 x + 18 y - 32

  Fx'(x,y,1).Fy'(x,y,1) = 4 b" F(x,y,1)

<=>
  (6 x - 2 y - 2)(-2 x + 9 y - 16) = -2(6 x2  - 4 x y + 9 y2  - 4 x - 32 y - 6)

<=>
        ...
<=>
        5 x y - 10 x - 5 y + 2 = 0  (this is a Plucker hyperbola )
Now we'll calculate the axes.
The center-point is (1,2,1).
The slopes m of the axes are the solutions of
 
        b" + (a' - a) m - b" m2  = 0
<=>

        -2 m2 - 3 m + 2 = 0
<=>
        m = 1/2 or m = -2

first axis: y - 2 = (1/2)(x - 1)  <=>  y = (1/2) x + 3/2
second axis: y - 2 = -2 (x - 1) <=>  y = -2 x + 4
The intersection points of the first axis and the Plucker hyperbola are the foci
 
     4  ___      2  ___              4  ___      2  ___
(1 - - V 5 , 2 - - V 5 )  and   (1 + - V 5 , 2 + - V 5 )
     5           5                   5           5

The intersection points of the second axis and the Plucker hyperbola are not real.

Equation of a conic section with given focus, directrix and eccentricity

Given :
Point F(xo,yo) is a focus.
Line d with equation u x + v y + w = 0 is the corresponding directrix.
e is the eccentricity.
 
        Point P(x,y) is on the conic section
<=>
        |P,F|2  = e2 .|P,d|2
<=>

                2           2     e2
        (xo - x)  + (yo - y)  = --------(u x + v y + w)2
                                u2 + v2

The last equation is the equation of the conic section.

Polar equation of a not degenerated conic section.

Circle

See polar equation of a circle .

Conic section K.

Choose the pole in a focus of the conic section and the polar axis orthogonal with the directrix of that focus. Choose an y-axis through the pole and orthogonal with the polar axis. The x-axis and the y-axis form an orthonormal cartesian basis.

The cartesian equation of the directrix is x + k = 0.
From above we know that the cartesian equation of the conic section is

 
    x2 + y2 = e2 (x + k)2
The corresponding polar equation is
 
    r2 = e2 (r cos(t) + k)2

<=>
    r = e (r cos(t) + k)   or   r = - e (r cos(t) + k)
<=>
    r(1 - e cos(t)) = ke   or   r(1 + e cos(t)) = - ke
<=>
           ke                             - ke
    r = ---------------  (1)  or   r = -------------      (2)
        1 - e cos(t)                   1 + e cos(t)
You can verify that :
If point d(r,t) is a solution of (1) then it is a solution of (2).
If point d(r,t) is a solution of (2) then it is a solution of (1).

From this it follows that (1) and (2) are polar equations of the same conic section K.

A polar equation of a not degenerated conic section different from a circle is

 
            k e                           - ke
    r = ------------        or     r = -------------
        1 - e cos(t)                   1 + e cos(t)
If t = pi/2 then r= ke. Let p = ke. Then we have

A polar equation of a not degenerated conic section different from a circle is
 
             p                            - p
    r = ------------        or     r = -------------
        1 - e cos(t)                   1 + e cos(t)

To calculate intersection points of the conic section with another curve, it is useful to know if the polar equation has the property (P) ( see Common points of two curves ).

Not ALL polar coordinates of EACH point of the conic section are solutions of

 
             p
    r = ------------
        1 - e cos(t)
But ALL polar coordinates of EACH point of the conic section are solutions of
 
             p                            - p
    r = ------------        or     r = -------------
        1 - e cos(t)                   1 + e cos(t)

and this is equivalent with

    r(1 - e cos(t)) = p  or   r(1 + e cos(t)) = - p

<=>
    r(1 - e cos(t)) = ke  or   r(1 + e cos(t)) = - ke

<=>
    r = e (r cos(t) + k)   or   r = - e (r cos(t) + k)
<=>
    r2 = e2 (r cos(t) + k)2
The equation r2 = e2 (r cos(t) + k)2 of the conic section has the property (P).

The equation r2 = e2 (r cos(t) + k)2 of the conic section is equivalent with the cartesian equation x2 + y2 = e2 (x + k)2.




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