Parabola Equation

Equation of a parabola in its standard form

AS = AK

Distance of A from the focus = Distance A from the directrix

⇒ A lies on the parabola

Let SK = 2a

then AS = AK = a Since SK = AS + AK

Choose A = (0,0) the origin.

AS is the x-axis, and let AY perpendicular AS be the y-axis ( stands for perpendicular)

Coordinates of S = (0,0). Equation of the directrix is a line parallel to the y-axis at a distance of “a” from it.

∴ x = -a since ZZ1 is to the left of the y-axis

Let P be a point on the parabola.

Join SP. Draw PM and PN perpendicular to ZZ1 and the x-axis, respectively.

Then

PM = NK

= AK + AN

= x+a

Now P lies on the parabola.

  1. Symmetry:
    For every positive value of x, there are two equal and opposite values of y. So the curve is symmetric about the positive part of the x-axis to the right of the y-axis.

  2. Region:
    For every negative value of x, y is imaginary. Therefore, no points of the curve lie to the left of the y-axis.

  3. Origin:
    The curve passes through the origin.

  4. Intersection with the axes:
    The curve meets the coordinate axes only at the origin.

The distance of a point P(x, y) from the focus S is called the focal distance of the point P.

S = (a, 0)

= x + a

The other standard forms of the parabola are

y2 = - 4ax

x2 = 4ay

x2 = -4ay

The different forms of these parabolas are drawn below

  1. y2= 4ax

  2. y2= -4ax

  3. x2 = 4ay

  4. x2 = -4ay

For these four standard forms, certain corresponding results are given below.

 

y2 = 4ax

y2 = -4ax

x2 = 4ay

x2 = -4ay

Coordinates of the Vertex

(0,0)

(0,0)

(0,0)

(0,0)

Coordinates of the Focus

(a,0)

(-a,0)

(0,a)

(0,-a)

Equation of the Directrix

x = -a

x = a

y = -a

y = a

Equation of the Axis

y = 0

y = 0

x = 0

x = 0

Length of the Latus Rectum

4a

4a

4a

4a

Focal Distance of a point P(x,Y)

x + a

a - x

y + a

y – a

Focus, directrix, latus rectum, axis for a given parabola

Example 3:

For the following parabolas find the coordinates of the foci, equations of directorices and the lengths of the latus rectum.

  1. y2 = 8x
  2. x2 = 6y
  3. y2 = -12x
  4. x2 = - 16y




To find the vertex, focus, axis, directrix, latus rectum of parabolas reducible to one of the four standard forms

Example 4:

Find the vertex, axis, focus, directrix and latus rectum of the following parabolas.

  1. y2 - 8y - x + 19 = 0
  2. 4y2 + 12x - 20y + 67 = 0
  3. y = x2 - 2x + 3

Shifting the origin to the point (3, 4)

we denote the coordinates with respect to new axes by X and Y.

             x - 3 = X              y - 4 = Y

      then x = X + 3             y = Y + 4

where (x,y) represent points with respect to the original axes.

Equation (1) becomes

Vertex:

Vertex:

With respect to the new axes, the vertex is (X = 0, Y = 0) or (0,0)

With respect to the old axes

             x = 0 + 3 y = 0 + 4

              x = 3,     y = 4

         or (3, 4)

Axis:

With respect to the new axes

            Y = 0

With respect to the old axes

             y = Y + 4

             y = 0 + 4

             y = 4

Focus:

With respect to the new axes

             (a, 0) = (1/4, 0)

With respect to the old axes

             x = X + 3

                = 1/4 + 3

                = 13/4

             y = Y + 4

                = 0 + 4

                = 4

          Focus (13/4, 4)

Directrix:

With respect to the new axes

             X = -a

       ⇒   X = -1/4

With respect to the old axes

             x = X + 3

                = -1/4 + 3

                = 11/4

Latus rectum:

4a = 1

We can draw the curve as follows.

Shifting the origin to the point (-7/2, 5/2) denoting the new coordinates with respect to these axes by X and Y.

             x + 7/2 = X, y-5/2 = Y

             x = X - 7/2, y = Y + 5/2

Using these relations, Equation (1) reduces to

             y2 = - 3X

This is of the form Y2 = - 4aX

             - 4a = -3

                 a = 3/4

Vertex:

Coordinates of the vertex with respect to the new axes are (0, 0)

With respect to old axes

             x = X - 7/2

                = 0 - 7/2

                = -7/2

             y = Y + 5/2

                = 0 + 5/2

                = 5/2

        Vertex (-7/2, 5/2)

Axis:

With respect to the new axes

                Y = 0

With respect to old axes

                y = Y + 5/2

                   = 0 + 5/2

                y = 5/2

      or 2y - 5 = 0

Focus:

With respect to the new axes (-a, 0)

With respect to the old axes

             x = X - 7/2

                = -3/4 - 7/2

=   

               = -17/4

             y = Y + 5/2

             y = 0 + 5/2

              y = 5/2

          Focus (-17/4, 5/2)

Directrix:

With respect to the new axes

              X = a = 3/4

With respect to the old axes

           x = X - 7/2

              = 3/4 - 7/2

=   

              = -11/4

Latus rectum:

4a = 3

The curve is given below.

4a = 1

     ∴     a = 1/4

Vertex:

With respect to the new axes

     X = 0, Y = 0

With respect to the old axes

     x = 0 + 1      y = 0 + 2

     x = 1            y = 2

     Vertex = (1, 2)

Focus:

With respect to the new axe

          (0, a) = (0, 1/4)

With respect to the old axes

           x = 0 + 1

              = 1

           y = 1/4 + 2

              = 9/4

Focus (1, 9/4)

Axis:

With respect to the new axes, X = 0

With respect to the old axes

     x = X + 1

       = 0 + 1

     x = 1

  or x-1 = 0

Directrix:

With respect to the new axes

           Y = -a = -1/4

With respect to the old axes

           y = Y + 2

              = -1/4 + 2

            y = 7/4

Latus rectum:

4a = 1

The figure of the parabola is given below..

Try these questions

I.    Find the equation of the parabola whose

i.

Focus = (1, 1) Directrix is x + y + 1= 0.

ii.

Focus = (2, 3) Directrix is x - 4y + 3= 0

iii.

Focus = (-1, -2) Directrix is x - 2y + 1= 0


II.    Find the vertex, focus, axis, directrix and latus rectum of the following parabolas. Also, give a rough sketch of the curve.

i.

y2 = 12x

ii.

4x2 + y = 0

iii.

y2 - 4y - 3x + 1 = 0

iv.

x2 + y = 6x - 14

v.

4(y-1)2 = -7(x-3)

Answers

Solutions

  1. Let P(x,y) be a point on the parabola
    S = (1,1)
    Directrix = x + y + 1
    PM = length of the perpendicular from point P on the directrix

  2. Let P(x,y) be a point on the parabola
    Focus S = (2,3)
    Directrix = x - 4y + 3 = 0
    PM = length of the perpendicular from point P on the directrix

  3. Let P(x,y) be a point on the parabola whose focus S = (-1, -2)
    and directrix = x - 2y + 1= 0
    PM = the perpendicular from point P on to the directrix




  1. Vertex:

    With respect to the new axes (X, Y) = (0, 0)

    With respect to the old axes

                     x = 0 - 1 = -1

                     y = 0 + 2 = 2

                  Vertex (-1, 2)

    Axis:

    With respect to the new axes Y = 0

    With respect to the old axes

                     y = Y + 2

                        = 0 + 2

                      y = 2

                or y-2 = 0

    Focus:

    With respect to the new axes

                  focus = (a, 0)=7:12 AM 9/19/2004(3/4, 0)

    With respect to the old axes

                     x = X - 1

                        = 3/4 - 1

                        = 3-4/4

                        = -1/4

                     y = Y + 2

                        = 0 + 2

                        = 2

    Focus (-1/4, 2)

    Directrix:

    With respect to the new axes

                     X = -a

                        = -3/4

    With respect to the old axes

                     x = X - 1

                       = -3/4 - 1

                       = -3-4/4

                     x = -7/4

          or 4x + 7 = 0

    Latus Rectum:

    4a = 3

    The figure is as follows



  2. Vertex:

    With respect to the new axes (X, Y) = (0, 0)

    With respect to the old axes

                                       x = x + 3

                                          = 0 + 3

                                          = 3

                                       y = Y - 5

                                          = 0 - 5

                                          = -5

                                    Vertex (3, -5)

    Axis:

    With respect to the new axes X = 0

    With respect to the old axes

                                       x = X + 3

                                          = 0 + 3

                                       x = 3

                                or x-3 = 0

    Focus:

    With respect to the new axes

    Focus S = (0,-a) = (0,-1/4)

    With respect to the old axes

                                       x = X + 3

                                          = 0 + 3

                                          = 3

                                       y = Y - 5

                                          = -1/4 - 5

                                          = -1-20/4

                                          = -21/4

                               Focus S = (3, -21/4)

    Directrix:

    With respect to the new axes

                                       Y = +a

                                          = 1/4

    With respect to the old axes

                                       y = Y - 5

                                          = 1/4 - 5

                                          = 1-20/4

                                        y = -19/4

                             or 4y + 19= 0

    Latus Rectum:

    4a = 1

    The figure is as follows

  3.     

                 x = 0 + 3

                        = 3

                     y = 0 + 1

                        = 1

    Vertex (3,1)

    Axis:

    With respect to the new axes Y = 0

    With respect to the old axes

                     y = 0 + 1

                        = 1

             or y - 1 = 0

    Focus:

    With respect to the new axes

               = (-a, 0), (-7/16, 0)

    With respect to the old axes

                     x = X + 3

                        = -7/16 + 3

    =   

                        = 41/16

                     y = Y + 1

                        = 0 + 1

                        = 1

    Focus (41/16, 1)

    Directrix:

    With respect to the new axes

                     X = a

                        = 7/16

                     x = X + 3

                        = 7/16 + 3

    =   

    x = 55/16

    or 16x - 55 = 0

    Latus Rectum:

    4a = 7/4

    The figure is as follows