Equation of a parabola in its standard form |
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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. |
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- 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.
- Region:
For every negative value of x, y is imaginary. Therefore, no points of the curve lie to the left of the y-axis.
- Origin:
The curve passes through the origin.
- 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. |
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S = (a, 0) |
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= x + a
The other standard forms of the parabola are |
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y2 = - 4ax
x2 = 4ay
x2 = -4ay |
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The different forms of these parabolas are drawn below |
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| i. |
y2= 4ax |
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| ii. |
y2= -4ax |
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| iii. |
x2 = 4ay |
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| iv |
x2 = -4ay |
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For these four standard forms, certain corresponding results
are given below. |
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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 |
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Focus, directrix, latus rectum, axis for a given parabola |
Example 3: |
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For the following parabolas find the coordinates of the foci, equations of directorices and the lengths of the latus rectum. |
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- y2 = 8x
- x2 = 6y
- y2 = -12x
- x2 = - 16y
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To find the vertex, focus, axis, directrix, latus rectum of parabolas reducible to one of the four standard forms |
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Example 4: |
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Find the vertex, axis, focus, directrix and latus rectum of the following parabolas. |
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- y2 - 8y - x + 19 = 0
- 4y2 + 12x - 20y + 67 = 0
- y = x2 - 2x + 3
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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 |
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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. |
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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 |
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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 |
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= -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 |
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= -11/4
Latus rectum:
4a = 3
The curve is given below. |
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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.. |
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Try these questions |
I. Find the equation of the parabola whose |
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Focus = (1, 1) Directrix is x + y + 1= 0. |
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Focus = (2, 3) Directrix is x - 4y + 3= 0 |
iii. |
Focus = (-1, -2) Directrix is x - 2y + 1= 0 |
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II. Find the vertex, focus, axis, directrix and latus rectum of the following parabolas. Also, give a rough sketch of the curve. |
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y2 = 12x |
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4x2 + y = 0 |
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y2 - 4y - 3x + 1 = 0 |
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x2 + y = 6x - 14 |
v. |
4(y-1)2 = -7(x-3) |
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Answers |
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Solutions |
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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 |
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ii. |
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 |
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iii. |
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 |
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II |
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ii. |
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iii. |
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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 |
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iv. |
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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 |
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v. |
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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 |
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= 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 |
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x = 55/16
or 16x - 55 = 0
Latus Rectum:
4a = 7/4
The figure is as follows |
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