The Parabola




Definition and equation

Take a line d and a point F not on d.
The locus of all points P such that |P,d| = |P,F| is a parabola.
To obtain an equation, we choose the x-axis and y-axis as in the figure below.
 
             
We give F coordinates (p/2,0).
Then we have d with equation x = - p/2.
 
        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
The point F is called the focus and the line d is the directrix. The axis of symmetry of the parabola is called the axis of the parabola. The axis of the parabola is the line through the focus and perpendicular to the directrix. The intersection point of that axis and the parabola is called the vertex of the parabola.

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(xo, a xo2) of the parabola the slope of the tangent line is 2 a xo.

  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"ll show that all parabolas are similar.

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".
Thus all parabolas are similar to the same fixed parabola.
This means that all parabolas are similar.

Congruent parabolas

The shape of the parabola is determined by a geometric property. It does not depends on the chosen coordinate system.

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
Example:
The parabola P1 with equation y = 0.25 x2 is congruent with the parabola P2 with directrix d2 : x - y = 0.
The focus of P2 is on the line b with equation x + y - 1 = 0.
Find all possible equations of P2.

The focus of P1 is (0,1) and the directrix is y+1=0. This focus lies at a distance 2 from the directrix.

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)
So there are two possible locations for the F2.
  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

Take in a plane two lines a and b with resp. equations
 
        x = 2 p t2             (1)

        y = 2 p t               (2)
The real number t is the parameter.
We know, from the theory of 'Elimination of parameters', that the intersection points of the two associated lines constitute a curve. To obtain the equation of that curve, we eliminate the parameter t from the two equations. This means that we search for the condition such that (1) and (2) has a solution for t.
From (2) we have t = y / (2p).
So, this t-value is a solution of (1) if and only if
 
                 y  2
       x = 2 p (---)    <=>  y2  = 2 p x
                2 p

Hence, the two associated lines constitute a curve and that curve is the parabola.
We say that (1) and (2) are parametric equations of the parabola. The point
 
D( 2 p t2  , 2 p t)
is on the parabola for each t-value and with each point of the parabola corresponds a t-value.

Tangent line in a point D of a parabola

Take the parabola
 
             y2  = 2p x
To obtain the slope of the tangent line we differentiate implicitly.
 
        2 y y' = 2 p
<=>
        y' = p/y
Say D(xo,yo) is a fixed point of the parabola.
The slope of the tangent line in point D is
 
         p
        ---
         y0
The equation of the tangent line is
 
               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)
The last equation is the tangent line in point D(x0,y0) of a parabola.
It is easy to show that this line meets the x-axis at the point s(-x0,0).
From this it is easy to construct the tangent line in a given point D. (see figure)
 
    

Powerful properties

From this we deduce many properties.

Tangent line with a given slope

Take a line t with a given slope m. The equation is y = m x + q.
The intersection points with the parabola are the solutions of the system
 
        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

Take a fixed point P(x0,y0) .
We'll calculate the tangent lines from P to the parabola .
A line t with variable slope through P is
 
        y - y0 = m(x - x0)
The intersection points with the parabola are the solutions of the system
 
        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
The line t is a tangent line if and only if the roots of the last equation (in y) are equal. Therefore the discriminant has to be zero.
 
        4 p2  + 4 m (2 p m x0 - 2 p y0) = 0
<=>
        2 x0 m2  - 2 y0 m + p = 0
The roots of this equation are the slopes of the two tangent lines.
 

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

                  y0 - sqrt(y02 - 2 p x0)
           m2 =  ----------------------------
                          2 x0
The equations of the tangent lines are
 
                     y2
        y - y0= m1(----- - x0)       ;
                     p

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

The two lines are orthogonal if and only if m1 . m2 = -1
 
         p
<=>     ---- = -1
        2 x0


               -p
<=>     x0 = ----
               2
From this we see that if point P is on the directrix, the tangent lines are orthogonal.

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

See 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.




Topics and Problems

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