Problems about Analytic Geometry

• READ THIS FIRST
  
• About imaginary points and lines
  
• About degenerated conic sections
  
• About a tangent lines
  
• About asymptotes
  
• About systems of conic sections
  
• About pole and polar line
  
• About center-point of a conic section
  
• About center-line of a conic section
  
• About axes of an affine conic section
  
• About foci of a conic section
  
• About the locus of a point
  

READ THIS FIRST

About imaginary points and lines

• 
  

  Are the following lines k and l conjugate imaginary lines? line k has line coordinates (1-i,i,4) line l has line coordinates (2,-1-i,4-4i)

  
  

   
          The lines k and l are conjugate imaginary lines
  <=>
          The lines (1+i,-i,4) and (2,-1-i,4-4i) are coinciding
  <=>
          (1+i,-i,4) and (2,-1-i,4-4i) are proportional
  

  Since (2,-1-i,4-4i) = (1-i).(1+i,-i,4), the lines k and l are conjugate imaginary lines.

• 
  

  Calculate three imaginary points on the line x - 2 y + z = 0

  
  

  Choose two real points (2,1,0) and (1,0,-1).
  (2,1,0) + r.i.(1,0,-1) are imaginary points on the line.
  For r = 1, 2, 3 we have the points
  (2 + i, 1, -i), (2 + 2 i, 1, -2 i), (2 + 3 i, 1, -3 i)

• 
  

  Calculate the real point on the line k(2 + i, 1, -i)

  
  

  The real point is the intersection point of the lines k(2 + i, 1, -i) and l(2 - i, 1, i). This point is ( i, -4 i, 2 i) or (1, -2, 1)

Level 2 problems

• 
  

  Calculate the real values m and n such that the point (2 - i, 3 - n i, m + i) is a real point.

  
  

   
          (2 - i, 3 - n i, m + i) is a real point
  <=>
          There are real values b and c such that
          3 - n i = b (2 - i)  and
          m + i   = c (2 - i)
  <=>
          b = 3/2; n = 3/2; m = -2; c = -1
  
  Then the point is
  
          (2 - i, 3 - 3/2 i, -2 + i)
  <=>
          (2 - i, (3/2)(2 - i), -1(2 - i))
  <=>
          (1, 3/2, -1)
  <=>
          (2, 3, -2)
  

About degenerated conic sections

• 
  

  Calculate the components of the curve x2 + 4 x - 6 = 0

  
  

   
          x2 + 4 x - 6 = 0
  <=>
               ____                 ____
          x = V 10  - 2  or  x = - V 10  - 2
  

  These are two lines parallel to the y-axis.

• 
  

  Calculate the components of the curve x2 + 4 x y + 3 y2 - 2 x z - 4 y z + z2= 0

  
  

  To factor this expression, we consider it as a quadratic equation of x. Collecting terms involving x, we have:

   
          x2 + (4 y - 2 z) x + (3 y2 - 4 z y + z2) = 0
  
  The discriminant is
          (4 y - 2 z)2 - 4 (3 y2 - 4 z y + z2)
  
          = 4 y2
  
  The roots are
  
          x = z - y, and  x = z - 3 y
  
  The two components are
  
          x + y - z = 0  and x + 3 y - z = 0
  

• 
  

  Calculate the real values of m such that the following conic section is degenerated. x2 - 4 x y + 2 y2 - 5 y - m x + 2 = 0

  
  

  The homogeneous equation is

   
          x2 - 4 x y + 2 y2 - 5 y z - m x z + 2 z2= 0
  
          The conic section is degenerated.
  <=>
          DELTA = 0
  <=>
          |   1      -2     -m/2 |
          |   -2      2     -5/2 | = 0
          | -m/2    -5/2      2  |
  <=>
          2 m2 + 20 m + 41 = 0
  <=>
              3  ___               3  ___
          m = - V 2  - 5  or m = - - V 2  - 5
              2                    2
  

• 
  

  Calculate the double points of the following conic section x2 + 4 x y + 3 y2 - 2 x z - 4 y z + z2= 0

  
  

   
  The double point is the solution of the system
          2 x + 4 y - 2 z = 0
          4 x + 6 y - 4 z = 0
          -2 x - 4 y + 2 z = 0
  The solution is
          x = 1, y = 0, z = 1
  

• 
  

  Calculate the double points of the following conic section x2 + 2 x y + y2 - 8 x z - 8 y z + 16 z2= 0

  
  

   
  The double point is the solution of the system
          x + y - 4 z = 0
          x + y - 4 z = 0
          -4 x - 4 y + 16 z = 0
  
  The system is equivalent with the single equation
  
          x + y - 4 z = 0
  

  All the solutions of this equation are double points.
  The conic section is a degenerated parabola. It consists of two coinciding lines. The equation can be written as

   
          (x + y - 4 z)2 = 0
  

About a tangent lines

• 
  

  Calculate the tangent line in point P(2,0) of the conic section x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0

  
  

   
  
  The equation is
  
          2(2 x - 4 y + 2 z)+0(-4 x - 2 y - 4 z)+1(2 x - 4 y - 16 z) = 0
  <=>
          x - 2 y - 2 z = 0
  

• 
  

  Calculate the tangent line in point P(2,0) of the conic section x2 + x y - 6 y2 - 4 x z - 2 y z + 4 z2= 0

  
  

   
  
  The formula gives
  
          2(2 x + y - 4 z) + 0(x - 12 y - 2 z) + 1(-4 x - 2 y + 8 z) =0
  <=>
          0 = 0
  this is not the equation of a line.
  

  Since F_x'(2,0,1) = 0 and F_y'(2,0,1) = 0 and F_z'(2,0,1) = 0 the point P is a double point and each line through P is a tangent line.

• 
  

  The tangent line in point P(?,?) of the conic section x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0 is x - 2 y - 2 = 0 Calculate the coordinates of the tangent point.

  
  
  • 
    method 1.
    The conic section and the line intersect in the tangent point. The coordinates are the solution of the system

     
            / x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0
            \ x - 2 y - 2 = 0
    

    Substituting x from the second equation in the first one, we find y = 0 and from this x = 2. Point P is (2,0).

  • 
    method 2.
    The tangent point is the intersection point of the tangent line and the tangent chord of a simple point on the tangent line. Take Q(0,-1) on the tangent line.
    The tangent chord is -x + 2 y + 2 = 0.
    The tangent point is the solution of the system

     
            / -x + 2 y + 2 = 0
            \ x - 2 y - 2 = 0
    

    The solution is P(2,0).
  • 
    method 3.
    Say P(x_o,y_o,1) is the tangent point. The tangent line in P has equation

     
            x(2 xo - 4 yo + 2)+y(-4 xo - 2 yo - 4)+z(2 xo - 4 yo - 16)=0
    

    That line is the same line as x - 2 y - 2 = 0. The corresponding coefficients must be directly proportional.
    So, x_o and y_o are the solutions of the system

     
            / 2 xo - 4 yo + 2 = r
            | -4 xo - 2 yo - 4 = -2r
            \ 2 xo - 4 yo - 16 = -2r
    

    The solution is x_o = 2, y_o = 0, r = 6.

About asymptotes

• 
  

  Calculate the asymptotes of the conic section x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0

  
  
  The slopes m corresponding with the ideal points are the solutions of

   
          - m2 - 8 m - 5  = 0
  <=>
                ____                     ____
          m1 = V 11  - 4  and m2 = -4 - V 11
  
  The asymptotes are the tangent lines in the ideal points
   
                 ____                         ____
          P1(1, V 11  - 4,0) and  P2(1, -4 - V 11  ,0)
  
  The asymptotes are
                                ____
          (2 x - 4 y + 2 z) + (V 11  - 4)(-4 x - 2 y - 4 z) = 0
  and
                                 ____
          (2 x - 4 y + 2 z) + (-V 11  - 4)(-4 x - 2 y - 4 z) = 0
  

• 
  

  Calculate the asymptotes of the conic section x2 + 2 x y + y2 - 4 x - 5 y + 7 = 0

  
  
  Since delta = 0, the conic section has two coinciding ideal points.
  Since DELTA is not zero, that ideal point can't be a double point.
  Therefore, the tangent line in that ideal point is the ideal line. The asymptotes are z = 0 and z = 0.
• 
  

  Calculate the asymptotes of the conic section x2 + 2 x y - 4 x = 0

  
  
  This conic section is degenerated in the lines

   
          x = 0 and x + 2 y - 4 = 0 .
  

  These lines are the asymptotes because these lines are the tangent lines in the ideal points of te conic section.
• 
  

  Calculate the asymptotes of the conic section x2- 2 x = 0

  
  
  The conic section is a degenerated parabola.
  The ideal point (0,1,0) is a double point of the degenerated conic section. Each line through this point is an asymptote. So this conic section has an infinity number of asymptotes. All these asymptotes are parallel.
• 
  

  Search the equation of the conic section with asymptotes x = 0 and x + 2 y - 41 = 0 and through the point P(2,1).

  
  

   
  
  The  conic section has an equation of the form
  
          x (x + 2 y - 41) + k = 0
  
  The point P(2,1) is on the conic section if and only if
  
          2 (2 + 2 - 41) + k = 0
  <=>
          k = 74
  
  The conic section has equation
  
          x (x + 2 y - 41) + 74 = 0
  <=>
          x2 + 2 x y - 41 x + 74 = 0
  

About systems of conic sections

• 
  

  Calculate the equation of the system of conic sections with basic points A(1,2); B(2,0); C(-1,1); D(0,3).

  
  

   
  Line AB:        2 x + y - 4 z = 0
  Line CD:        -2 x + y - 3 z = 0
  

  The lines AB and CD form a degenerated conic section of the system. We choose it as the first basic conic section. The equation of this conic section is

   
          (2 x + y - 4 z)(-2 x + y - 3 z) = 0
  <=>
          -4 x2 + y2 + 2 x z - 7 y z + 12 z2= 0
  
  
  Line AC:        x - 2 y + 3 z = 0
  Line BD:        -3 x - 2 y + 6 z = 0
  

  The lines AC and BD form a degenerated conic section of the system. We choose it as the second basic conic section. The equation of this conic section is

   
          -3 x2 + 4 x y + 4 y2 - 3 x z - 18 y z + 18 z2= 0
  

  The equation of the system of conic sections with basic points A(1,2); B(2,0); C(-1,1); D(0,3) is

   
          k(-4 x2 + y2 + 2 x z - 7 y z + 12 z2)
                  + l( -3 x2 + 4 x y + 4 y2 - 3 x z - 18 y z + 18 z2) = 0
  

  k and l are homogeneous parameters.
  With non-homogeneous parameter h, we have the equation:

   
           (-4 x2 + y2 + 2 x z - 7 y z + 12 z2)
                  + h( -3 x2 + 4 x y + 4 y2 - 3 x z - 18 y z + 18 z2) = 0
  

• 
  

  Calculate the equation of the system of conic sections with basic points A(1,2); B(2,0); C(-1,1); C(-1,1) and such that the line c with equation x + y = 0 is the tangent line in point C.

  
  

  Degenerated conic sections of the system are line AB with line c, and AC with line BC. The equations of these degenerated conic sections are

   
          (2 x + y - 4 z)(x + y) = 0
  <=>
          2 x2 + 3 x y + y2 - 4 x z - 4 y z = 0
  
  and
          (x - 2 y + 3 z)(-x - 3 y + 2 z) = 0
  <=>
          - x2 - x y + 6 y2 - x z - 13 y z + 6 z2= 0
  
  

  The system has equation:

   
          k(2 x2 + 3 x y + y2 - 4 x z - 4 y z)
                  + l(- x2 - x y + 6 y2 - x z - 13 y z + 6 z2) = 0
  

• 
  

  Calculate the equation of the system of conic sections with basic points A(1,2); A(1,2); B(2,0); B(2,0) and such that the line a with equation x + y - 3 z = 0 is the tangent line in point A and b with equation x + 2 y - 2 z = 0 is the tangent line in point B.

  
  

  Degenerated conic sections of the system are line a with line b, and line AB with line AB. The equations of these degenerated conic sections are

   
          (x + y - 3 z)(x + 2 y - 2 z) = 0
  and
          (2 x + y - 4 z)2= 0
  

  The system has equation:

   
          k(x + y - 3 z)(x + 2 y - 2 z) + l(2 x + y - 4 z)2= 0
  

• 
  

  Calculate the basic points of the system with basic conic sections x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 and -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2= 0

  
  

  The basic points are the solutions of the system

   
          /
          | x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0
          |
          | -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2= 0
          \
  

  To calculate these points we first calculate a value of r such that

   
          (x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2)
            + r(-3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2) = 0
  

  is a degenerated conic section.

   
                  | 2 - 6 r   2 - 4 r    3 r - 5  |
          DELTA = | 2 - 4 r  14 + 10 r  -17 - 9 r |
                  | 3 r - 5  -17 - 9 r  12 + 12 r |
  
                = -300 (r + 1) (r - 1)2
  

  The conic section is degenerated for r = -1 and r = 1.
  We take r = 1. Then the degenerated conic section is

   
          -2 x2 - 2 x y + 12 y2 - 2 x z - 26 y z + 12 z2= 0
  <=>
          (x - 2y + 3 z)(x +3 y - 2 z) = 0
  

  The basic points are the solutions of the systems

   
          /
          | (x - 2y + 3 z) = 0
          |
          | -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2
          \
  and
  
          /
          | (x +3 y - 2 z) = 0
          |
          | -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2
          \
  
  

  The first system has solutions (-1,1,1) and (1,2,1).
  The second system has solutions (2,0,1) and (-1,1,1).
  These points are the four basic points of the system.

About pole and polar line

• 
  

  Calculate the polar line of P(1,1,1) relative to the conic section -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2= 0

  
  

   
  
  The polar line is
  
          1.Fx' (x,y,z) + 1.Fy' (x,y,z) + 1.Fz' (x,y,z) = 0
  <=>
          -7 x - 3 y + 6 z = 0
  

• 
  

  Calculate the polar line of P(1,1,1) relative to the conic section x2 - y2 - 2 x z + 2 y z = 0

  
  

   
          1.Fx' (x,y,z) + 1.Fy' (x,y,z) + 1.Fz' (x,y,z) = 0
  <=>
          2 x - 2 z + 2 z - 2 y + 2 y - 2 x = 0
  <=>
          0 = 0
  

  This method gives no result because the point P(1,1,1) is a double point of the conic section. Each line is a polar line of a double point.

• 
  

  Calculate the polar line p of P(1,0,1) relative to the conic section (x - y) (x + y - 2 z) = 0 This conic section has double point S(1,1,1). Show that the components of the conic section, the line SP and the line p form a harmonic quartet of lines.

  
  

  The polar line is

   
          1.Fx' (x,y,z) + 0.Fy' (x,y,z) + 1.Fz' (x,y,z) = 0
  <=>
          2 y - 2 z = 0
  <=>
          y - z = 0
  

  The line SP is x - z = 0.

  The equations of the four lines can be written as

   
          x - z = 0
          y - z = 0
          (x - z) - 1.(y - z) = 0       (parameter h = 1)
          (x - z) + 1.(y - z) = 0       (parameter h' = - 1)
  

  Since h = -h' , we have a harmonic quartet of lines.

• 
  

  Calculate the point C of a polar triangle ABC of the conic section x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 if you know that A(2,1,1) and B(0,15,1).

  
  
  The point C is the intersection point of the polar lines of A and B. So, it is the solution of the system

   
          / x + y - 15 z = 0
          |
          \ 25 x + 193 y - 243 z = 0
  

  The coordinates of C are (221,-11,14).
• 
  

  The lines a,b and c are the polar lines of the vertices of a triangle ABC, relative to a not degenerated conic section K. P is the intersection point of a and BC. Q is the intersection point of b and CA. R is the intersection point of c and AB. Prove that P,Q and R are collinear.

  
  
  Since all properties in this problem are projective, we can solve it in the projective plane.

   
  Choose A(1,0,0) B(0,1,0) and C(0,0,1)
  
  Let K : a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2= 0
  
  Then
  
  line a has equation  a x + b" y + b' z = 0
  line b has equation  b" x + a' y + b z = 0
  line c has equation  b' x + b y + a" z = 0
  
  and
  
  P(0,b',-b")  Q(b,0,-b") R(b,-b',0)
  
  Since
           | 0     b'     -b"|
           | b     0      -b"|  = 0
           | b    -b'      0 |
  the points P, Q and R are collinear.
  
  
  

About center-point of a conic section

• 
  

  Calculate the center-point of the conic section x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0

  
  
  The coordinates of the center-points are the solutions of the system

   
           / Fx' (x,y,z) = 0
           \ Fy' (x,y,z) = 0
  <=>
           / 2 x + 2 y - 5 z = 0
           \ 2 x + 14 y - 17 z = 0
  <=>
          x = 3 ; y = 2 ; z = 2
  

  The center point is (3,2,2).

• 
  

  Calculate the center-point of the conic section x2 + 4 x y + 4 y2 + 2 x z + 4 y z - 8 z2 = 0

  
  
  The coordinates of the center-points are the solutions of the system

   
           / Fx' (x,y,z) = 0
           \ Fy' (x,y,z) = 0
  <=>
           / 2 x + 4 y + 2 z = 0
           \ 4 x + 8 y + 4 z = 0
  

  All the points of the line x + 2 y + z = 0 are center-points of the degenerated parabola.

• 
  

  What is the general equation of a conic section with the point (0,0,1) as a center-point.

  
  
  A conic section has an equation of the form

   
  a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2= 0
  

  The coordinates of the center-points are the solutions of the system

   
           / Fx' (x,y,z) = 0
           \ Fy' (x,y,z) = 0
  <=>
           / a x + b" y + b' z = 0
           \ b" x + a' y + b z = 0
  
          (0,0,1) is a solution of this system
  <=>
          b' = b = 0
  

  The general equation of a conic section with the point (0,0,1) as a center-point is

   
          a x2 + 2 b" x y + a' y2 + a" z2= 0
  

About center-line of a conic section

• 
  

  Calculate the center-line of the conic section x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 conjugated to the direction with slope -1.

  
  
  This center line is the polar line of the point (1,-1,0).
  This line has equation

   
          Fx' (x,y,z) - Fy' (x,y,z) = 0
  <=>
          2 x + 2 y - 5 z - (2 x + 14 y - 17 z) = 0
  <=>
          z - y = 0
  

  So, the center-line is the line y = 1.

• 
  

  Calculate the value of r such that the line x + 2 y + z = 0 is a center-line of the conic section x2 + 2 x y - y2 - r x z + r y z = 0

  
  

   
  The center-line of the point (1,m,0) is
  
          Fx' (x,y,z) + m  Fy' (x,y,z) = 0
  <=>
          2 x + 2 y - r z + (2 x - 2 y + r z) m = 0
  <=>
          (2 m + 2) x + (2 - 2 m) y + (r m - r) z = 0
  
          This line is the line x + 2 y + z = 0
  <=>
          (2 m + 2)    (2 - 2 m)     (r m - r)
          --------- = ----------- = -----------
              1           2               1
  <=>
                1
          m = - -, r = -1
                3
  

• 
  

  Calculate the value of r and s such that the line x + 2 y + z = 0 is a center-line conjugated to the direction with slope -2 relative to the conic section x2 + 2 x y - y2 - r x z + s y z -2 z2 = 0

  
  

   
  The center-line of the point (1,-2,0) is
  
          Fx' (x,y,z) - 2  Fy' (x,y,z) = 0
  <=>
          2 x + 2 y - r z - 2 (2 x - 2 y + s z) = 0
  <=>
          -2 x + 6 y - (2 s + r) z = 0
  
  We see that there are no values of r and s such that the line
  x + 2 y + z = 0  coincide with the line -2 x + 6 y - (2 s + r) z = 0
  

• 
  

  Calculate the direction conjugated to (1,-2,0) relative to the conic section x2 + 2 x y - y2 - 4 x z + 2 y z -2 z2 = 0

  
  

   
          (1,-2,0) and (1,m,0) are conjugated directions
  <=>
          a r1 r2 + b"(r1 s2 + s1 r2) + a' s1 s2 = 0
  <=>
          1.1.1  +  1.(1 .m  - 2 . 1) - 1 .(-2).m = 0
  <=>
          m = 1/3
  

  The directions (1,-2,0) and (1,1/3,0) are conjugated.

• 
  

  Calculate the center-line conjugated to the center-line 4 x - 2 y + z = 0 relative to the conic section x2 + 2 x y - y2 - 4 x z + 2 y z -2 z2 = 0

  
  
  The center-line conjugated to the center-line 4 x - 2 y + z = 0, is the center line conjugated to the direction of 4 x - 2 y + z = 0. So, we calculate the center-line conjugated to (1,2,0).
  This line is 3 x - y = 0.

About axes of an affine conic section

• 
  

  Calculate the axes and the vertices of the conic section x2 + 2 x y - (1/2) y2 - 4 x z + 2 y z -6 z2 = 0

  
  
  First we calculate the main directions

   
          (1,m,0) is a main direction
  <=>
          b" + (a' - a) m - b" m2 = 0
  <=>
          1  + (-1/2  - 1) m - 1  m2 = 0
  <=>
          m = -2 ; m = 1/2
  

  The axes are the center-lines conjugated to these directions.

   
          2 y - x - 4 z = 0
  and
          2 x + y - 2 z = 0
  

  The vertices on the axis 2 y - x - 4 z = 0 are the solutions of the system

   
          / 2 y - x - 4 z = 0
          |
          \ x2 + 2 x y - (1/2) y2 - 4 x z + 2 y z -6 z2 = 0
  
  We choose z = 1
          / 2 y - x - 4  = 0
          |
          | x2 + 2 x y - (1/2) y2 - 4 x + 2 y - 6 = 0
          \
  
  
  The vertices are  the points
             4   ___       2   ____
          (- -- V 30 , 2 - -- V 30  )
             15            15
  
           4   ___       2   ____
          (-- V 30 , 2 + -- V 30  )
           15            15
  

  Similarly, you'll find the vertices on the other axis

• 
  

  Calculate the axis of the parabola x2 + 2 x y + y2 - 4 x + 2 y - 6 = 0

  
  
  The ideal point of the parabola is (1,-1,0). The orthogonal direction is (1,1,0). The axis of the parabola is the polar line of (1,1,0).
  It is the line 2 x + 2 y - 1 = 0.

About foci of a conic section

• 
  

  Calculate the focus and the directrix of the parabola x2 + 2 x y + y2 - 4 x + 2 y - 6 = 0

  
  
  The focus of the parabola is the intersection point of the two Plucker lines. These Plucker lines have equation

   
          /
          |  (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)
          \
  
  <=>
          / 2 y + 2 x - 1 = 0
          |
          \ 3 x - 3 y + 4 = 0
  <=>
                 5       11
           x = - --, y = --
                 12      12
  
          The focus is (-5,11,12)
  

  The directrix is the polar line of the focus. This directrix is

   
          -5 Fx' (x,y,1) + 11 Fy' (x,y,1) + 12 Fz' (x,y,1) = 0
  <=>
          -6 x + 6 y + -17 z = 0
  

• 
  

  The origin point (0,0,1) is the focus of a not degenerated parabola. The distance from the origin to the vertex of the parabola is r. Show that the distance from the origin to the directrix is 2r.

  
  
  First method:
  Choose the x-axis on the axis of the parabola such that the coordinates of the vertex are (-r,0,1). Now the equation of the parabola is

   
          y2 = 4 r (x + r)
  <=>
          y2 = 4 r (x z + r z2)
  <=>
          y2 - 4 r x z - 4 r2 z2= 0
  
  The directrix is the polar line of the origin and has equation
  
          -4 r x - 8 r2 z = 0
  
  <=>
          x + 2 r z = 0
  <=>
          x = - 2 r
  

  It is clear that the distance from the origin to this directrix is 2r.

  Second method:
  Since the vertex of each parabola is the midpoint between the focus and the directrix and since the origin is the focus, the distance from the origin to the directrix is 2r.

About the locus of a point

• 
  

  Given is a quadratic equation in z with parameters x and y. z2 - x z + (x - y)2 = 0 The parameters x and y are the coordinates of a point P relative to an orthonormal coordinate system in a plane. Calculate the locus of point P such that the quadratic equation has equal roots.

  
  

   
       The quadratic equation has equal roots
  
  <=>  The discriminant D = 0
  
  <=>  x2 - 4(x - y)2 = 0
  
  <=>  -3 x2 + 8 x y - 4 y2 = 0
  
  

  This is the equation of the locus of P. It is the quadratic equation of two lines through the origin of the coordinate system. The slopes of the lines are 1/2 and 3/2.

• 
  

  A variable circle c has equation x2 + y2 - 2 (t2 - 3 t + 1) x - 2 (t2 + 2 t) y + t = 0 The number t is a parameter. Calculate the locus of the center of the circle.

  
  
  The center of the circle is C(t² - 3 t + 1, t² + 2 t). The center is the intersection point of the lines

   
    x = t2 - 3 t + 1   and y = t2 + 2 t
  

  We eliminate t and we find the equation of a parabola.

   
        x2 - 2 x y + y2 - 12 x - 13 y + 11 = 0
  

• 
  

  We have, in an orthonormal coordinate system, a circle c through the origin O and with a fixed radius R. Now, the circle c starts to rotate about the origin O. From a fixed point at infinity, we draw tangent lines at the rotating circle. Calculate the locus of all points of tangency.

  
  
  In most problems the difficulty of calculations depends highly on the choice of the coordinate system.

  Here we choose the coordinate system such that the given point at infinity is (1,0,0).

  The variable center point of c has coordinates (R cos(t), R sin(t)). The number t is the parameter.

  The equation of the variable circle is

   
  (x - R cos(t))2 + (y - R sin(t))2 = R2      (1)
  

  The points of tangency are on the tangent chord. It is the polar line of point (1,0,0). It is also the line through the center point of the circle orthogonal to the x-axis. The equation of that line is

   
  x = R cos(t)             (2)
  

  (1) and (2) are the associated curves. We eliminate the parameter t from this two equations.

   
  (2) in (1) gives
  
        (y - R sin(t))2 = R2
  
  <=>  y - R sin(t) = R  or y - R sin(t) = -R
  
  
  <=>  R sin(t) = y - R   or  R sin(t) = y + R    (3)
  

  We have to eliminate t from (2) and (3).

   
  (2)2 + (3)2 gives
  
        x2 + (y - R)2 = R2  or x2 + (y + R)2 = R2
  

  The locus consists of two circles with center points (0,R) and (0,-R) and radius R.
• 
  

  Calculate the locus of a point P such that the tangent lines from P to the ellipse b2x2 + a2y2 - a2b2 = 0 are orthogonal lines.

  
  
  The quadratic equation of the tangent lines through point P(x₁,y₁,1) has equation:

   
  (x.Fx' (x1,y1,1) + y.Fy' (x1,y1,1) + Fz' (x1,y1,1))2
                  - 4 F(x,y,1).F(x1,y1,1) = 0
  
  For the ellipse, this equation becomes:
  
  (x.2b2 x1 + y.2a2 y1 -2a2 b2)2
           -4(b2 x12 + a2 y12 - a2 b2)(b2 x2 + a2 y2 - a2 b2) = 0
  
  The tangent lines are orthogonal lines if and only if the sum of the
  coefficients of x2 and y2 is 0.
  This condition is here
  
     4b4 x12 - 4(b2 x12 + a2 y12 - a2 b2)b2 + 4 a4 y12
            -4((b2 x12 + a2 y12 - a2 b2)a2 = 0
  
  <=> -a2 b2 y12 + a2 b4 -a2 b2 x12 + a4 b2 = 0
  
  <=>  x12 + y12 = a2 + b2
  
  

  The equation of the locus is the circle x² + y² = a² + b²

• 
  

  The conic section with equation m x2 + 2 xy + m y2 + 2m x - 2m y = 0 is variable because of the parameter m. Find the locus of the variable center-point.

  
  

   
      P(xo,yo) is center-point
  <=>
      Fx' (xo,yo,zo) = 0 and Fy' (xo,yo,zo) = 0
  <=>
      P(xo,yo) is intersection point of the lines
  
      Fx' (x,y,z) = 0 and  Fy' (x,y,z) = 0
  
  <=>
      P(xo,yo) is intersection point of the lines
  
      m x + y + m = 0 and x + m y - m = 0
  

  The equation of the locus arises if we eliminate the non homogeneous parameter m.
  We find :

   
      x2 - y2 + x + y = 0
  
  <=>
      (x + y)(x - y + 1) = 0
  

  The locus consists of two lines.