1-The Parabola
Why study the parabola?
The parabola has many applications in situations where:
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Radiation needs to be concentrated at one point (e.g. radio
telescopes, pay TV dishes, solar radiation collectors) or
Radiation needs to be transmitted from a single point into a wide
parallel beam (e.g. headlight reflectors).
Here is an animation showing how parallel radio waves are collected
by a parabolic antenna.
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Definition of a Parabola
The parabola is defined as the locus of a point which moves so that it
is always the same distance from a fixed point (called the focus) and
a given line (called the directrix).
In the following graph,
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The focus of the parabola is at (0, p).
The directrix is the line y = -p.
The focal distance is |p| (Distance from the origin to the focus,
and from the origin to the directrix. We take absolute value
because distance is positive.)
The point (x, y) represents any point on the curve.
The distance d from any point (x, y) to the focus (0, p) is the
same as the distance from (x, y) to the directrix.
[The word locus means the set of points satisfying a given condition.
See some background in Distance from a Point to a Line.]
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The Formula for a Parabola - Vertical Axis
Adding to our diagram from above, we see that the distance d = y + p.
Now, using the distance formula on the general points (0, p) and (x, y),
and equating it to our value d = y + p, we have
Squaring both sides gives:
(x − 0)2 + (y − p)2 = (y + p)2
Simplifying gives us the formula for a parabola:
x2 = 4py
In more familiar form, with "y = " on the left, we can write this as:
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where p is the focal distance of the parabola.
Example - Parabola with Vertical Axis
Sketch the parabola
Find the focal length and indicate the focus and the directrix on
your graph.
The focal length is found by equating the general expression for y
and our particular example:
So we have:
This gives p = 0.5.
So the focus will be at (0, 0.5) and the directrix is the line y = -0.5.
Our curve is as follows:
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Note: Even though the sides look as though they become straight as x
increases, in fact they do not. The sides of a parabola just get steeper
and steeper (but are never vertical, either).
Arch Bridges − Almost Parabolic
The Gladesville Bridge in Sydney, Australia was the longest single
span concrete arched bridge in the world when it was constructed in
1964.
The shape of the arch is almost parabolic, as you can see in this
image with a superimposed graph of y = −x2 (The negative means the
legs of the parabola face downwards.)
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[Actually, such bridges are normally in the shape of a catenary, but that is
beyond the scope of this chapter.]
Parabolas with Horizontal Axis
We can also have the situation where the axis of the parabola is
horizontal:
In this case, we have the relation:
y2 = 4px
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[In a relation, there are two or more values of y for each value of x. On the other
hand, a function only has one value of y for each value of x.]
Example - Parabola with Horizontal Axis
Sketch the curve and find the equation of the parabola with focus (2,0) and directrix x = 2.
In this case, we have the following graph:
After sketching, we can see that the equation required is in the
following form, since we have a horizontal axis:
y2 = 4px
Since p = -2 (from the question), we can directly write the equation of
the parabola:
y2 = -8x
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Shifting the Vertex of a Parabola from the Origin
This is a similar concept to the case when we shifted the centre of a
circle from the origin.
To shift the vertex of a parabola from (0, 0) to (h, k), each x in the
equation becomes (x − h) and each y becomes (y − k).
So if the axis of a parabola is vertical, and the vertex is at (h, k), we
have
(x − h)2 = 4p(y − k)
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Exercises:
1. Sketch x2 = 14y
Solution
x2 = 14y
Focus:
4p = 14;
so
p = 14/4 = 3.5
So the focus is at (0, 3.5)
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Directrix: y = -3.5
2. Find the equation of the parabola having vertex (0,0), axis along the
x-axis and passing through (2,-1).
Solution:
The curve must have this orientation, since we know it has horizontal
axis and passes through (2, -1):
So we need to use the general formula
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y2 = 4px
We need to find p. We know the curve goes through (2, -1), so we
substitute:
(-1)2 = 4(p)(2)
→ 1 = 8p
→ p = 1/8.
So the required equation is y2 = x/2.
3. We found above that the equation of the parabola with vertex (h, k)
and axis parallel to the y-axis is
(x − h)2 = 4p(y − k).
Sketch the parabola for which (h, k) is (-1,2) and p = -3.
Answer
We will have vertex at (-1,2) and p = -3 (so the parabola will be
"upside down").
The vertex is at (-1, 2), since we know the focal distance is | p | = 3.
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We don't really need to find the equation, but as an exercise:
Using
(x − h)2 = 4p(y − k)
We have:
(x + 1)2 = 4(-3)(y − 2)
x2 + 2x + 1 = -12y + 24
12y = -x2 − 2x − 1 + 24
y = (-x2 − 2x + 23)/12
Application:
A parabolic antenna has a cross-section of width 12 m and depth of 2
m. Where should the receiver be placed for best reception?
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The receiver should be placed at the focus of the parabolic dish for best reception,
because the incoming signal will be concentrated at the focus.
We place the vertex of the parabola at the origin (for convenience) and use
the equation of the parabola to get the focal distance ( p) and hence the
required point.
In general, the equation for a parabola with vertical axis is
x2 = 4py.
We can see that the parabola passes through the point (6, 2).
Substituting, we have:
(6)2 = 4p(2)
So p = 36/8 = 4.5
So we need to place the receiver 4.5 metres from the vertex, along the axis of
symmetry of the parabola.
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The equation of the parabola is:
x2 = 18y
That is
y = x2 /18)
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