The Parabola

• Definition and equation
  
• Other forms of the equation of a parabola
  
• Similar parabolas
  
• Congruent parabolas
  
• Parametric equations of the parabola
  
• Tangent line in a point D of a parabola
  
• Powerful properties
  
• Tangent line with a given slope
  
• Tangent lines from a given point
  
• Evolute of a parabola
  
• Exercises:
  

Definition and equation

 
        P(x,y) is on the parabola

                <=>

            |P,d| = |P,F|

                <=>

            |P,d|2 = |P,F|2

                <=>

            p 2        p 2
       (x + -)  = (x - -)  + y2
            2          2
                <=>

                ...

                <=>

             y2  = 2p x

Other forms of the equation of a parabola

1. We swap the x-axis and y-axis

   Then we have a parabola with an equation

    
       x2 = 2 p y
   <=>
              1
       y = ----- x2
            2 p
   
     Now the focus is  (0, p/2) and the directrix is  y = -p/2.
   
     Let a = 1/ (2 p)
   
     Now the equation is
   
       y = a x2
   
   
                
   
   

   The focus is ( 0, 1/(4a) ) and the directrix is y = - 1/(4a)

   Since the equation is y = ax² , we can find the slope of the tangent line by means of the derivative. In point P(x_o, a x_o²) of the parabola the slope of the tangent line is 2 a x_o.

2. A translation of the parabola y = a x² gives a parabola with an equation of the form
   y = ax² + bx + c

3. The equation y² = 2p x owes its simplicity to the special location of the directrix and the focus relative to the axes.

   However, the parabola was defined as a set of points which satisfy a geometric condition. Now, we set up the equation of a parabola with focus F(a,b) and directrix ux+vy+w=0.

   Point P(x,y) is on the parabola if and only if |P,d| = |P,F|.
   Since we need the distance from the line ux+vy+w=0 , we first bring this equation in its normal form
   l x + m y + n = 0 with l² + m² =1.

    
           P(x,y) is on the parabola
   
                   <=>
   
               |P,F| = |P,d|
   
                   <=>
   
               |P,F|2 = |P,d|2
   
                   <=>
   
       (x - a)2 + (y - b)2 = (l x + m y + n)2               (1)
   
   

   Example:

   We calculate the equation of the parabola with focus F(1,2) and directrix d: 3x+4y-2=0.
   The directrix has normal equation (3/5) x + (4/5)y - 2/5 = 0.
   The equation of the parabola is

    
   (x-1)2 + (y-2)2 = ((3/5) x + (4/5)y - 2/5)2             (2)
   

   It is obvious that this form can be simplified to

    
   16 x2 - 24 xy + 9 y2 - 38 x - 84 y + 121 = 0              (3)
   

   The annoying thing is that that form not shows where the focus is and where te directrix is. If we want F and d, we have to calculate F en d from the last expression. Therefore we bring (1) in another shape.

    
       (x - a)2 + (y - b)2 = (l x + m y + n)2
   <=>
       x2 - 2ax + a2 + y2 -2by + b2 = l2 x2 + m2 y2 + n2  + 2 l.m xy + 2 l.n x + 2m.n y
   <=>
       (1-l2)x2 -  2 lm xy + (1-m2) y2 = 2x(a + l.n) + 2y(b + m.n) -a2 - b2 + n2
   <=>
         (m x - l y)2 = 2x(a + l.n) + 2y(b + m.n) -a2 - b2 + n2    (4)
   

   We transform (3) to the shape (2).

    
        16 x2 - 24 xy + 9 y2 - 38 x - 84 y + 121 = 0
   <=>
        (4 x - 3 y)2 =  38 x + 84 y - 121
   
            Since l2 + m2 must be equal to 1, we divide both sides by 25
   
   <=>
         ( (4/5) x - (3/5) y)2 =  38/25 x + 84/25  y - 121/25
   
   We compare this result with (4) and we see that  m = (4/5) en l = (3/5)
   
        a + l.n = 19/25                      (5)
        b + m.n = 42/25                      (6)
        n2 - a2 - b2 = - 121/25             (7)
   
   Since l and m is known, we bring a and b from (5) and (6) to (7) and then
     a = 1 ; b = 2 ; n = -2/5.
   
   The equation (3) is a parabola with focus F(1,2) and directrix
     (3/5)x + (4/5)y -2/5 = 0  <=>   3x+4y-2=0.
   

4. Focus in O(0,0)
   x² +y² = (l x + m y + n)²

5. Focus in O(0,0) and the directrix parallel to the y-axis.
   x² +y² = (x-c)²
   Consider c as a parameter. Then we have a set of parabolas with a fixed focus and a fixed axis. Such a set of parabolas is called confocal parabolas.

6. Focus in O(0,0) and the directrix parallel to the x-axis.
   x² +y² = (y-c)²
   Again we have a set confocal parabolas. The parameter is c.

Similar parabolas

We start with a fixed coordinate system, an arbitrary parabola P and a fixed parabola P" with equation y = x².

There is always a suitable rotation and translation such that P is transformed in a parabola P' with equation y = ax² with a > 0.
So, P is similar to P'. We also know that a homothetic transformation transforms a figure in a similar figure. We choose O(0,0) as center of a homothety h and we choose 'a' as factor.
The transformation formulas are

 
           h
   (x,y) ----> (ax, ay)

   Point D' is on parabola P'

<=>    D'(x , a x2)

   h transforms this point D' in the point D"

         D"( a (x) , a (a x2)) = D"( (a x) , (a x)2 )

<=> Point D" is on the parabola P" with equation  y = x2

So, the arbitrary parabola P is similar to the fixed parabola P".

Congruent parabolas

A parabola is completely determined by a given focus and directrix.

Consider two parabolas :

 
   P1 with focus F1 and directrix d1
and
   P2 with focus F2 and directrix d2


     The parabolas P1 and P2 are congruent

<=>

   There is a displacement v such that   v(F1) = F2 and v(d1) = d2

<=>

   The relative position of F1 against d1
       is the same as
   The relative position of F2 against d2

<=>

    The distance from  F1 to d1 = the distance from F2 to d2

<=>

    The distance from  F1 to the tangent at the vertex of P1
             is equal to
    The distance from  F2 to the tangent at the vertex of P2

The parabola P₂ is congruent with P₁ if and only if the focus F₂ lies at a distance 2 from the directrix x - y = 0.

Let F₂ be a variable point of the line b. We use parameter t.
F₂ = F₂(t , 1-t).

 
       | F2, d2| = 2
<=>
        | 2 t -1|
       ------------- = 2
          sqrt(2)
<=>
        (2t - 1)2 = 8
<=>
      t = 0.5 ± sqrt(2)

1. t = 0.5 + sqrt(2)

    
       F2 = F2(0.5 + sqrt(2) , 0.5 -sqrt(2) )
   
       The equation of the parabola P2 is
   
       (x - 0.5 -sqrt(2))2  + (y - 0.5 + sqrt(2))2 = (x-y)2/2
   
   <=>
   
       x2 + 2 x y + y2 - 7.66 x + 3.66 y + 9 = 0
   
   
              
   
   
   

2. t = 0.5 - sqrt(2)

   Solve this case as exercise

Parametric equations of the parabola

 
        x = 2 p t2             (1)

        y = 2 p t               (2)

 
                 y  2
       x = 2 p (---)    <=>  y2  = 2 p x
                2 p

 
D( 2 p t2  , 2 p t)

Tangent line in a point D of a parabola

 
             y2  = 2p x

 
        2 y y' = 2 p
<=>
        y' = p/y

 
         p
        ---
         y0

 
               p
      y - y0 = --  (x - x0)
               y0
<=>
     y0 y - y02  = p x - p x0

                Since   y02  = 2p x0

<=>
     y0 y - 2 p x0 = p x - p x0
<=>
        y y0 = p (x + x0)

Powerful properties

• |C,D| = |D,F| = |E,F| = x₀ + p/2 and so CDEF is a rhomb.
  
• Hence the tangent line bisects the angle CDF.
  
• Point C is the mirror image of F relative to the tangent line.
  
• So, the mirror image of F relative to a variable tangent line is the directrix.
  
• Additionally, the orthogonal projection of F on a variable tangent line is the tangent line through the vertex of the parabola.
  
• The line through point D and orthogonal with the tangent line is called the normal at point D.
  The normal through D is also a bisecting line of CD and DF. D₁ = D₂.
  From this property it follows that parabolas have a certain reflection property. All light rays entering parallel to the axis of a parabola are reflected to the focus. This works in the other direction too. A point source placed at the focus will send light out in a collimated beam parallel to the axis.

Tangent line with a given slope

 
        y2  = 2 p x

        y = m x + q

Substitution gives
        (m x + q)2  = 2 p x
<=>
        m2  x2  + 2 (m q - p) x + q2  = 0

The line t is a tangent line if and only if the roots of the last
equation are equal. Therefore the discriminant has to be zero.

        4 (m q - p)2  - 4 m2  q = 0
<=>
        4 p (p - 2 m q) = 0
<=>
             p
        q = ---
            2 m
The tangent line with a given slope m is

                       p
        y =     m x + ---
                      2 m

Tangent lines from a given point

 
        y - y0 = m(x - x0)

 
        y2  = 2 p x

        y - y0 = m(x - x0)

We substitute x from the first equation into the second one.
                   y2
        y - y0 = m(--- - x0)
                   2 p
<=>
        - m y2  + 2 p y - 2 p y0 + 2 p m x0 = 0

 
        4 p2  + 4 m (2 p m x0 - 2 p y0) = 0
<=>
        2 x0 m2  - 2 y0 m + p = 0

 

                  y0 + sqrt(y02 - 2 p x0)
           m1 = ------------------------------
                          2 x0

                  y0 - sqrt(y02 - 2 p x0)
           m2 =  ----------------------------
                          2 x0

 
                     y2
        y - y0= m1(----- - x0)       ;
                     p

                      y2
         y - y0= m2(----- - x0)
                      p

 
         p
<=>     ---- = -1
        2 x0


               -p
<=>     x0 = ----
               2

The directrix is the locus of the points P such that the tangent lines through P to the parabola are perpendicular.

Evolute of a parabola

Exercises:

The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the problem before you read the solution.

• 
  

  Find the points of the parabola y2 = 4x such that the distance from these points to the focus is four.

  
  

  • First method

    The focus is F(1,0).
    The required points are the intersections of the parabola with a circle (x-1)² + y² = 16.
    We solve the system formed by the equation of the parabola and the equation of the circle. We find two points : (3, ± sqrt(12)).

  • Second method

    The points at a distance 4 from the focus, are also at a distance 4 of the directive. The directive has an equation x = -1. The required points are on the line x = 3.
    So they have an abscissa 3. We find (3, ± sqrt(12)).

• 
  

  The parabola P has equation y2 = 2 p x. A variable point P has coordinates (p/2,t). The parameter is the real number t greater than p . Calculate the the tangent of the sharp angle between the tangent lines through P.

  
  

  From the theory about the tangent lines through a given point P(x_o,y_o), we know that the slopes are given by

   
                           ______________
                          |   2
                    yo + \| yo  - 2 p xo
             m1 = ---------------------
                            2 xo
  
                           ______________
                          |   2
                    yo - \| yo  - 2 p xo
             m2 = ---------------------
                            2 xo
  

  In our case here, xo = p/2 and yo = t. So, the slopes are

   
                           ______________
                          |   2    2
                    t  + \|  t  - p
             m1 = ---------------------
                            p
  
                           ______________
                          |   2    2
                    t  - \|  t  - p
             m2 = ---------------------
                            p
  
  The tangent of the angle between the lines is given by
  
           m1 - m2
          -----------
          1 + m1 m2
  Now,
                          __________
                          |   2    2
                       2 \|  t  - p
          m1 - m2  =  --------------   and  m1.m2 = 1
                           p
  Then the tangent of the angle between the lines is
  
                      __________
                     |   2    2
                    \|  t  - p
                   -------------
                       p
  

• 
  

  The tangent lines t and t' in points P(x1,y1) and in P'(x2,y2) of a parabola are orthogonal. Prove that y1.y2 = - p.p

  
  

  From the theory about the tangent line in a given point P(x_o,y_o) of a parabola, we know that the slope is p/y_o .
  Thus, in point P is the slope p/y₁ and in point P' is the slope p/y₂.
  The tangent lines are orthogonal if and only if

   
             p    p
             -- . -- = -1
             y1   y2
  
  <=>     y1.y2 = - p.p
  

• 
  

  Point P is on a parabola. The projection of P on the axis of the parabola is Q. The normal line in P intersects the axis in point N. Show that |QN| is constant.

  
  

  P( 2 p t² , 2 p t) is a variable point on the parabola. Then Q is Q( 2 p t² , 0).
  The tangent line in P has slope p/(2pt) = 1/(2t) .
  The slope of the normal line is -2t. The normal line in P is y - 2pt = -2t(x - 2pt²).
  The normal line in P intersects the x-axis in point N(p+2pt²,0). |QN| = p = constant.

• 
  

  The line y = x+3 is a tangent to the parabola y2 = 2px. Calculate p.

  
  

  The tangent line in a variable point P( 2 p t² , 2 p t) of the parabola is 2pt y = x + 2pt².
  That line should coincide with the given line. It follows that p = 1/6.

• 
  

  The line y = 0.5 x - 4 is a normal line to the parabola y2 = 2px. Find the tangent line to the parabola corresponding with that normal line.

  
  

  The tangent line is orthogonal to the normal line and so the slope is -2. The tangent line with that direction is y = -2 x - p/4

• 
  

  We consider two points P and P' on a parabola. The tangent lines in these points intersect on the directrix. Find the product of the ordinates of P and P'.

  
  

  Take P( 2 p t² , 2 p t) and P'( 2 p t'² , 2 p t') on the parabola. We calculate the tangent lines in P and P'. We find x - 2ty + 2pt² = 0 and x - 2t'y + 2pt'² = 0.
  We require that the two tangent lines and the directrix are concurrent lines. The condition is:

   
  | 1     -2t   2pt2   |
  | 1     -2t'  2pt'2  | = 0
  | 2      0      p    |
  
  <=>
  
   4t.t'= -1
  

  Then the product of the two ordinates is -p².

• 
  

  Consider the tangent lines in 4 points of a parabola. They intersect and form a quadrilateral. Show that the line through the midpoints of the diagonals of the quadrilateral is parallel to the axis of the parabola.

  
  

  For i = 1,2,3,4 is:

  We take 4 points P_i(2pt_i², 2pt_i) on the parabola.
  The four tangent lines l_i are x - 2 t_i y + 2 p t_i² = 0.
  The ordinate of the intersection point of l₁ and l₂ is p.(t₁ + t₂)
  So, the ordinate of the intersection point of l_i and l_j is p.(t_i + t_j) .

  We calculate the y-value of the center of each diagonal. We find (1/2).p.(t₁+t₂+t₃+t₄).

  Since the two centers have the same ordinate, the line through these centers is parallel to the axis of the parabola.

• 
  

  Take four points on a parabola. Find the condition in order that the four point are concyclic.

  
  

  For i = 1,2,3,4 is:

  We take 4 points P_i(2pt_i², 2pt_i) on the parabola.
  

  From the theory of the circle we know that the 4 points are concyclic if and only if

   
  | 4p2t14+4p2t12     2pt12     2pt1    1|
  | 4p2t24+4p2t22     2pt22     2pt2    1|  = 0
  | 4p2t34+4p2t32     2pt32     2pt3    1|
  | 4p2t44+4p2t42     2pt42     2pt4    1|
  

  We simplify this determinant as much as possible by canceling factors from the columns. Then we replace column1 by column1 - column2.

   
  |t14  t12   t1    1|
  |t24  t22   t2    1|  = 0
  |t34  t32   t3    1|
  |t44  t42   t4    1|
  

• 
  

  We know that x2 + y2 = (x - c)2 is the equation of a set confocal parabolas with focus F(0,0). Q(r,s) is a point different from F. Show that there are exactly two parabolas of the set through Q. Show that there is an orthogonal intersection of these two parabolas in point Q.

  
  

  Part 1:

  For each c-value there is a parabola in the set. We'll find the two c-values such that Q(r,s) is on a parabola.

   
        Q is on a parabola of the set
  <=>
        r2 + s2 = (r - c)2
  <=>
       ...
  <=>
       c2 - 2 r c - s2 = 0          (*)
  

  The discriminant of the quadratic equation is 4(r²+ s²). Since Q is different from F, D is > 0. So, there are two values for c such that Q is on the parabola. There are 2 parabolas through Q.

  Part 2:

  Say c₁ and c₂ are the two roots of (*). First, we take the parabola corresponding with c = c₁.

   
      x2 + y2 = (x - c1)2
  <=>
      y2 = - 2 c1 x + c12
  
      We calculate the derivative by  Implicit differentiation
  
      2 y y' = - 2 c1
  <=>
        y' = - c1/y
  

  In Q(r,s) the slope of the tangent line is - c₁/s
  In the same way we find for c = c₂.
  In Q(r,s) the slope of the tangent line is - c₂/s
  The product of these slopes is c₁.c₂/s²
  But from (*) we see that the product of the roots is c₁c₂ = -s².
  From this, the product of the slopes is c₁.c₂/s² = -1.
  So, the tangent lines are orthogonal in Q.
• 
  

  Consider a variable tangent line to the parabola y2 = 2 p x. This line cuts the x-axis at point Q and the y-axis at point R. Find the locus of the center M of [QR].

  
  You can find here theory and examples about loci.

  A point P(x_o, y_o) is on a parabola.
  x_o and y_o are parameters connected by the binding equation y_o² = 2 p x_o.
  The tangent line in a random point P(x_o,y_o) is y y_o = p (x + x_o).
  The intersection points with the axes are Q(-x_o , 0) and R(0, px_o/y_o ).
  The center M is M(-x_o/2 , px_o/ (2y_o) ).
  M is the intersection point of the associated lines x = -x_o/2 and y = px_o/ (2y_o).
  To find the locus, we eliminate x_o and y_o between the associated lines and the binding equation.

   
   / x = -xo/2
   | y = pxo/ (2yo)
   \ yo2 = 2 p xo
  

  We calculate x_o and y_o from the first two equations and we put the results in the last one. The equation of the locus is y² = - p x/4.
• 
  

  A variable line l has a fixed slope m. The line l intersects the parabola y2 = 2px in the points Q and R. Show that the center M of [QR] lies on a fixed line parallel to the axis of the parabola.

  
  

  De line l has an equation y = mx + q. q is a parameter. The points Q and R are the solutions of the system

   
    / y = m x + q
    \ y2 = 2 p x
  <=>
    / x = (y-q)/m
    \ y2 = 2p (y-q)/m
  <=>
    / x = (y-q)/m
    \ y2 - 2py/m + 2pq/m = 0          (*)
  

  We call y₁ and y₂ the solutions of the quadratic equation (*). y₁ and y₂ are the ordinates of Q and R.
  The ordinate of the center M is (y₁ + y₂)/2 = half the sum of the roots of (*). So, the ordinate of the center M is p/m. This is a fixed value as it is independent of q. M is on a fixed line parallel to the axis of the parabola.

• 
  

  P is the parabola y2 = 2p x. We connect a variable point P of the parabola with the vertex of the parabola. Show that the locus of the center M of this variable chord PO is a parabola.

  
  

  The transformation which transforms P into M is actually a homothetic transformation h with center O and factor 1/2. The image of P through h is a figure that is similar to the parabola P. Therefore, this figure is a parabola.

• 
  

  A parabola has an equation x2 + y2 = (x+1)2. Find the focus F and the directrix. A variable line through F intersects the parabola in the points A and B. Find the locus of the center M of [AB].

  
  

  The focus is F(0,0) and the directrix is x = -1.
  The equation of the parabola can be simplified to y² = 2x + 1.
  The variable line through F is y = m x. m is parameter.
  The intersection points A and B are the solutions of the system

   
      y2 = 2x + 1    (1)
      y = m x        (2)
  
    If we replace y in (1) by m x, then we get m2 x2 - 2 x -1 = 0
    The solutions of this equation are the x-values x1 and x2 of A and B.
  
    The center of [AB] has coordinates
  
      x = (x1 + x2)/2  = (2/m2) / 2 = 1/m2
      y = m x = 1/m
  
    So,   M( 1/m2, 1/m )
  

  The coordinates of M tell us, without calculation, that M moves along the curve x=y². The locus of M is the parabola y²= x.

• 
  

  Given Parabola with equation y = 0.25 x2 Points A(7,0) and B(3/2,-1) line r with equation y = 2 x - 4 Find: Focus and directrix ; make a figure The two tangent lines through point A The tangent line perpendicular to the line r

  
  

  • The directrix is y = -1 and F(0,1)
  • It is clear that the first tangent line through A is y = 0.
    

    Now, we retake the general figure with many properties of the parabola.

    We see that: if point S is on the tangent line through the vertex then the second tangent line through S is perpendicular to SF.

    In our exercise, the second tangent line through A is perpendicular to AF.
    The slope of AF is -1/7. So, the required tangent line has a slope = 7. The second tangent line is y = 7(x - 7).

  • Since r and r' are two perpendicular tangent lines, they intersect on the directrix. The intersection point of r with the directrix is R(1.5 , -1).
    The tangent line through R and perpendicular to r is
    y + 1 = - 0.5 (x - 1.5) <=> y = -0.5 x - 0.25

• 
  

  A is a variable point of the parabola y2=6x and B(2,8) is a fixed point. Find A such that |AB| is minimum.

  
  

  We use the parametric equations of the parabola. Take A(6t²,6t) with t as parameter.

   
    |AB| is minimum if and only if  |AB|2 is minimum
  
    |AB|2 = (6t2 - 2)2 + (6t - 8)2
  
            = 36 t4 + 12 t2 - 96 t + 68
  
     |AB| is minimum if and only if the  derivative is zero.
  
   <=>     12 t3 + 2 t - 8 = 0
  

  There is no simple algebraic way to solve this equation. Using a plot (For example, with this utility) we find t = 0.81.

  If a very accurate value should be obtained, we can use a numerical package or an iteration method .

  With t = 0.81 corresponds A(3.94 ; 4.86)

  If |AB| is minimum then B is on the normal line to the parabola through A.
  The equation of this normal line is y = -1.62 x + 11.24

  Check these results on a graph! (For example, with this utility)

• 
  

  The points A and B are on the parabola y2 = 2p x. M is the center of the segment [A,B]. The lines r an r' are the tangent lines in A and B to the parabola. The line s is parallel to the axis of the parabola and contains M. Show that the lines r, r' and s are concurrent lines.

  
  

   
     Let  A(2pt2 , 2pt) ,  B(2pt'2 , 2pt')
  
     Then  M(... , p(t+t'))
  
     line r :    y 2pt = p ( x + 2pt2)   <=>  x - 2 t y + 2 p t2 = 0
     line r':    y 2pt' = p ( x + 2pt'2) <=>  x - 2 t' y + 2 p t'2 = 0
     line s :    y =  p(t+t')             <=> 0x + y - p (t+t') = 0
  
     The lines are  concurrent if and only if
  
     | 1   -2t    2pt2  |
     | 1   -2t'  2pt'2  | = 0
     | 0    1   -p(t+t')|
  
             Row 2 - Row 1
  <=>
     | 1   -2t          2pt2  |
     | 0  2(t-t')  2p(t'2-t2) | = 0
     | 0     1       -p(t+t') |
  
  <=>
      -2p(t2-t'2) - 2p(t'2-t2) = 0
  

• 
  

  Find t such that the following parabolas are congruent. P1 : y2 = t x with t > 0 P2 : (x - 1)2 + (y - 2)2 = (0.6 x + 0.8 y)2

  
  

  The parabola P₁ has focus F₁(t/4, 0) and directrix d₁: x = - t/4.
  The distance from F₁ to d₁ is t/2.

  The parabola P₂ has focus F₂(1,2) and directrix d₂: 0.6 x + 0.8 y = 0.
  The distance from F₂ to d₂ is 2.2.

  So, t = 4.4
  The parabola y² = 4.4 x² is congruent with (x - 1)² + (y - 2)² = (0.6 x + 0.8 y)².