## 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)

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. 2. 3. 4. 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 5. 6. 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 7. 8.

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 