Proof:
b^{2} x^{2} + a^{2} y^{2} - a^{2} b^{2} z^{2} = 0The quadratic equation of the tangent lines out of focus point F(c,0,1) is
(c b^{2} x - a^{2} b^{2} z)^{2} - 4 (b^{2} c^{2} - a^{2} b^{2} ) (b^{2} x^{2} + a^{2} y^{2} - a^{2} b^{2} z^{2} ) = 0 m is slope of a tangent line <=> (1,m,0) is on a tangent line <=> c^{2} b^{4} + b^{4} (b^{2} - a^{2} m^{2}) = 0 <=> ... <=> m^{2} = -1
y^{2} - 2 p x z = 0The 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 p^{2} x^{2} + 4 p^{2} m^{2} = 0 <=> m^{2} = -1
Proof:
The isotropic lines through P(x_{o},y_{o}) have slope i and -i.
The equations are
y - y_{o} = i (x - x_{o}) and y - y_{o} = -i (x - x_{o}) <=> y - y_{o} - i (x - x_{o}) = 0 and y - y_{o} + i (x - x_{o}) = 0The quadratic equation of these lines is
(y - y_{o} - i (x - x_{o}))(y - y_{o} + i (x - x_{o})) = 0 <=> (y - y_{o})^{2} + (x - x_{o})^{2} = 0This is the equation of a circle with center P.
Proof:
If the quadratic equation of two lines is a circle, then the equation is
(y - y_{o})^{2} + (x - x_{o})^{2} - r^{2} = 0But if this is the equation of two lines, the circle is degenerated. Thus
DELTA = 0 <=> ... <=> r = 0Then, the equation is
(y - y_{o})^{2} + (x - x_{o})^{2} = 0Factorizing, we find
(y - y_{o} - i (x - x_{o}))(y - y_{o} + i (x - x_{o})) = 0This is the equation of two lines through a regular point.
Proof:
The ellipse has equation
b^{2} x^{2} + a^{2} y^{2} - a^{2} b^{2} z^{2} = 0P(x_{o},y_{o},1) is a real regular point. The tangent lines through P have as equation
(x.F_{x}'(x_{o},y_{o},1) + y.F_{y}'(x_{o},y_{o},1) + z.F_{z}'(x_{o},y_{o},1))^{2} - 4 F(x,y,z).F(x_{o},y_{o},1) = 0 <=> (x b^{2} x_{o} + y a^{2} y_{o} + z(- a^{2} b^{2} )) - ( b^{2} x^{2} + a^{2} y^{2} - a^{2} b^{2} z^{2} )( b^{2} x_{o}^{2} + a^{2} y_{o}^{2} - a^{2} b^{2} z^{2} )=0 These tangent lines are the isotropic lines <=> Previous equation is a circle <=> / | (b^{2} x_{o}^{2} ) - b^{2} ( b^{2} x_{o}^{2} + a^{2} y_{o}^{2} - a^{2} b^{2} z^{2} ) | | = (a^{2} y_{o})^{2} - a^{2} ( b^{2} x_{o}^{2} + a^{2} y_{o}^{2} - a^{2} b^{2} z^{2} ) | | b^{2} x_{o} a^{2} y_{o} = 0 \The second condition gives x_{o} = 0 or y_{o} = 0.
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
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 = 0We see that the condition for harmonic conjugate lines is satisfied.
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) = 0The equation of the line c, harmonic conjugate to b, then is
(y + ix) - 1(y - ix) = 0 <=> x = 0So, b and c are orthogonal.
b^{2} x^{2} + a^{2} y^{2} - a^{2} b^{2} z^{2} = 0If we calculate the polar line of F(c,0), we find
a^{2} x = -- cWe call this line the directrix of F.
a^{2} x = - -- cWe call this line the directrix of F'.
b^{2} x^{2} - a^{2} y^{2} - a^{2} b^{2} z^{2} = 0If we calculate the polar line of F(c,0), we find
a^{2} x = -- cWe call this line the directrix of F.
a^{2} x = - -- cWe call this line the directrix of F'.
Proof:
2 2 a 2 2 |P,d| = (xo - --) and |P,F| = (x_{o} - c)^{2} + y_{o}^{2} c with b^{2} x_{o}^{2} + a^{2} y_{o}^{2} = a^{2} b^{2} Thus a^{2} |P,F|^{2} = a^{2} (x_{o} - c)^{2} + a^{2} b^{2} - b^{2} x_{o}^{2} = ... = (c x_{o} - a^{2} )^{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| aThe constant value e = c/a is called the eccentricity of the ellipse.
|P,F| |P,F'| c ----- = ----- = --- = constant and > 1 |P,d| |P,d| aThe constant value e = c/a is called the eccentricity of the hyperbola.
___ eccentricity = e = V 2
P(x_{o},y_{o},1) is focus of this conic section <=> The tangent lines out of P are isotropic <=> (x.F_{x}'(x_{o},y_{o},1) + y.F_{y}'(x_{o},y_{o},1) + z.F_{z}'(x_{o},y_{o},1)) - 4 F(x,y,z).F(x_{o},y_{o},1) = 0 are isotropic <=> (x_{o},y_{o}) is a solution of the system / | (F_{x}'(x,y,1))^{2} - 4 a F(x,y,1) = (F_{y}'(x,y,1))^{2} - 4 a'F(x,y,1) | | | F_{x}'(x,y,1).F_{y}'(x,y,1) - 4 b" F(x,y,1) = 0 \ <=> (x_{o},y_{o}) is a solution of the system / | (F_{x}'(x,y,1))^{2} - (F_{y}'(x,y,1))^{2} = 4(a - a') F(x,y,1) | | | F_{x}'(x,y,1).F_{y}'(x,y,1) = 4 b" F(x,y,1) \
F(x,y,1) = 6 x^{2} - 4 x y + 9 y^{2} - 4 x - 32 y - 6 = 0 F_{x}'(x,y,1) = 12 x - 4 y - 4 F_{y}'(x,y,1) = - 4 x + 18 y - 32 F_{x}'(x,y,1).F_{y}'(x,y,1) = 4 b" F(x,y,1) <=> (6 x - 2 y - 2)(-2 x + 9 y - 16) = -2(6 x^{2} - 4 x y + 9 y^{2} - 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.
b" + (a' - a) m - b" m^{2} = 0 <=> -2 m^{2} - 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 + 4The 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 5The intersection points of the second axis and the Plucker hyperbola are not real.
Point P(x,y) is on the conic section <=> |P,F|^{2} = e^{2} .|P,d|^{2} <=> 2 2 e^{2} (xo - x) + (yo - y) = --------(u x + v y + w)^{2} u^{2} + v^{2}The last equation is the equation of the conic section.
The cartesian equation of the directrix is x + k = 0.
From above we know that the cartesian equation of the conic section is
x^{2} + y^{2} = e^{2} (x + k)^{2}The corresponding polar equation is
r^{2} = e^{2} (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 :
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) |
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) <=> r^{2} = e^{2} (r cos(t) + k)^{2}
The equation r^{2} = e^{2} (r cos(t) + k)^{2} of the conic section has the property (P). |
The equation r^{2} = e^{2} (r cos(t) + k)^{2} of the conic section is equivalent with the cartesian equation x^{2} + y^{2} = e^{2} (x + k)^{2}. |