One might have come across images of a radio telescope that are based on the ground. Their structure is similar to a dish, other objects such as a TV dish are also of a similar shape. Convex and concave mirrors also follow the pattern of that shape. The shape of such objects is in the form of a parabola. You may have heard the term ‘parabolic dish’ quite often, which is used in telecommunications every now and then.
Another place where we can observe the parabolic curve is a suspension bridge, the hanging cable suspensions are in the shape of a parabola, and is in order to keep it stable. But what exactly is a parabola or a parabolic object? How can a parabola be rigorously and mathematically defined?
The Definition of a Parabola
The definition of a parabola is as follows:
A parabola is the set of all points in a plane that are equidistant from a fixed point and a fixed-line.
We assume a fixed point on the plane, which is known as the Focus of the parabola. We also assume a fixed line, which is referred to in the definition and is known as the Directrix of the parabola. By definition, the set of points equidistant from the focus and the directrix form a parabola. The line passing through the focus and perpendicular to the directrix is called the Axis of the parabola.
The equation of a standard parabola is given as follows:
$$x^2=4ay$$
Fig. 1. A parabola with its directrix, focus and vertex signified.
The Graph of a Parabola
Observe that every parabola is symmetric with respect to its axis, i.e. the part of the parabola on one side will be identical to the part of the parabola on the other side. Here, we will cover standard parabolas, which are centered at the origin and the vertex lies on either the x-axis or the y-axis.
Fig. 2. A parabola whose axis is y-axis and directrix is y=-p and focus is at (0,p).
In the above diagram, the parabola is represented by the curved line, it can be seen that the parabola is symmetric around the y-axis, and thus y-axis is considered the axis of this parabola. And \(F\) is the focus of the parabola, and it is important to remember that the focus of every parabola always lies on the axis of the parabola. The straight line parallel to the x-axis is the directrix of the parabola, and it lies at a distance \(p\) from the parabola, and it lies on the other side of the focus.
Thus, the directrix is represented by the equation \(y=-p\) where \(p\) is a real-valued constant. The focus also lies at the same distance from the vertex as that of the directrix; thus the focus has coordinates \((0,p)\).
Equation of a standard parabola
To find the equation of a standard parabola, suppose that it is centered at the origin, as shown in the diagram. Let \(P\) be an arbitrary point on the parabola with coordinates \((x,y)\) where \(x\) and \(y\) are the variables. \(P\) represents any point on the parabola and is not fixed.
According to the definition, the distance between point \(P\) and the focus is the same as the distance between point \(P\) and the directrix.
$$d=\sqrt{x^2+(y-p)^2}$$
Whereas the distance between \(P\) and the directrix is \(|y+p|\), the modulus sign signifies that the point that distance is always positive. By definition, these distances are equal, which gives:
$$\sqrt{x^2+(y-p)^2}=|y+p|$$
Squaring both sides, we obtain
$$x^2+(y-p)^2=|y+p|^2$$
$$x^2+(y-p)^2=(y+p)^2$$
Expanding both sides
$$x^2+y^2-2yp+p^2=y^2+2py+p^2$$
Upon further simplification
$$x^2=4py$$
which is the Formula of a Parabola whose vertex lies on the origin.
An equation of the parabola with focus \((0,p)\) and directrix \(y=-p\) is \(x^2=4py\).
If we express \(a=\dfrac{1}{4p}\), then the above equation becomes \(y=ax^2\). Notice that if \(p>0\) then the parabola opens upwards and if \(p<0\) then the parabola opens downwards, in which case the focus will be on the negative y-axis and the directrix will be in the upper half-plane. But what about the parabola, whose axis is along the x-axis?
Interchanging \(x\) and \(y\) in \(x^2=4py\), we get
$$y^2=4px$$
Which is the standard form of a parabola symmetrical to the x-axis instead of the y-axis. The focus, in this case, lies at \((p,0)\) i.e. on the positive x-axis. The directrix shifts to \(x=-p\), which is parallel to the y-axis and perpendicular to the x-axis.
Find the focus and directrix of the parabola \(y^2+8x=0\).
Solution:
Rearranging the terms to get \(y^2\) on one side and \(x\) on another, we get \(y^2=-8x\) and comparing it with the standard form \(y^2=4px\), we get
$$4p=-8$$
Which yields \(p=-2\).
Therefore, the focus of the parabola is at \(F(-2,0)\), and its directrix is \(x=2\).
Find an equation of the parabola that has its vertex at the origin with the axis of symmetry lying on the y-axis, and passes through the point \(P(3,-4)\). What are the focus and directrix of the parabola?
Solution:
The equation of the described parabola is given by \(y=ax^2\). To determine the value of \(a\) we use the condition that \(P(3,-4)\) lies on this parabola. This point will satisfy the equation, which gives \(-4=a(3)^2\) which implies that \(a=\dfrac{-4}{9}\).
Therefore, by substituting the value of \(a\) in the equation of the parabola, we get
$$y=\dfrac{-4}{9}x^2$$
To find the focus of the parabola, observe that it has the form \(F(0,p)\). The value of \(p\) can be obtained using the value of \(a\) as follows:
$$p=\dfrac{1}{4a}=\dfrac{-9}{16}$$
Therefore, the focus is \(F\left(0,-\dfrac{9}{16}\right)\). And its directrix will be \(y=-p\), which turns out to be \(y=\dfrac{9}{16}\).
Parametric representation of a parabola
A parameter is a kind of variable that is constructed in such a way that each variable, x and y, can be expressed individually in terms of the parameter. The parameter is not a part of the Cartesian plane, it is not innate to the parabola, but only a way to connect the two variables indirectly.
Let the parameter be \(t\) such that \(t\in \mathbb R\), the way \(t\) is incorporated into the equation is \(x=pt^2\) and \(y=2pt\).
It should be noted that none of these equations exist on their own, but only collectively. The two equations can be combined to eliminate \(t\) to get the equation of the parabola \(y^2=4px\).
Why bother with Parametric Form?
The advantage of breaking down such an equation into parametric equations gives us the flexibility of mathematically working with the individual equations. Furthermore, there is not a specific parametrization that fits an equation. One can come up with as many parameters as one pleases, as long as it satisfies the equation.
Just as a circle can be described using polar coordinates, a parabola can also be expressed using polar coordinates. Polar coordinates are nothing but parameters. Indeed, polar coordinates are a special case of parametric equations.
Reflective property of the Parabola
All the mathematical analysis we have discussed so far is applied to real-world applications in a wide array of topics.
Suppose a point \(P\) on the parabola with focus \(F\), let \(I\) be the tangent line to the parabola at \(P\). The reflective property states that the angle \(\theta_1\) that lies between the line segment \(PF\) and the tangent line \(I\) is equal to the angle \(\theta_2\). \(\theta_2\) is the angle between line \(I\) and the line passing through \(P\) and parallel to the x-axis. This property has many applications:
Fig. 3. A parabola with its reflective property.
As mentioned earlier, the reflector of a radio telescope has a shape that is obtained by rotating the parabola around its axis, i.e. a parabola in three dimensions. A radio wave coming from a distant part of the solar system or the galaxy may be assumed to be parallel to the axis of the parabola. And the above property of the parabola allows the wave to converge towards the focus.
Fig. 4. A cross section of a radio telescope and that of a headlight of a vehicle.
In the headlights of all vehicles, this reflective property is used. The light bulb is placed at the focus of the parabola. It is placed in such a way so that when the light emanates from the reflector, the rays will spread in a direction parallel to the axis of the parabola.
Vertex Form of a Parabola
For some analogy, recall that a circle has an equation that we call as the center form of a circle, where we represent a circle directly as the coordinates of its center, i.e. \((x-h)^2+(y-k)^2=r^2\).
Similarly, we also have a form for the parabola, known as the Vertex form of a Parabola, which represents the parabola directly by the coordinates of its center.
A parabola, whose center is given by \((h,k)\), is represented by the following equation:
$$y=a(x-h)^2+k$$
It can be seen that using only the coordinates of the center and a point on the parabola, we can get the equation of a parabola. We can also do the opposite; using a given equation of a parabola, we can convert it into the above form and then determine the coordinates of the vertex.
Note that the above equation represents the class of parabolas that have their major axis parallel to the y-axis. Replacing \(x\) with \(y\), we get an accompanying equation that represents parabolas aligned in the direction of the x-axis.
Find the equation of the parabola whose vertex is \((2,3)\) and the point \((1,4)\) lies on the parabola.
Solution:
The vertex is \((h,k)=(2,3)\), which can be substituted in the general form \(y=a(x-h)^2+k\):
$$y=a(x-2)^2+3$$
But we still don't know the value of the constant \(a\).
For that, we will use the latter part of the given data; the point \((1,4)\) lies on the parabola.
Substituting for the above point in our obtained equation, we can determine the value of \(a\):
$$4=a(1-2)^2+3$$
$$1=a(1)$$
$$\therefore a=1$$
Now, we can substitute back in \(a=1\) to get our equation of the parabola:
$$y=1(x-2)^2+3$$
$$\therefore y=x^2-4x+4+3$$
$$\therefore y=x^2-4x+7$$
Hence, the equation of the required parabola is \(y=x^2-4x+7\).
Examples on Parabolas
Find the focus and directrix of the parabola \(y^2-16x=0\).
Solution:
Rearranging the terms to get \(y^2\) on one side and \(x\) on another, we get \(y^2=16x\) and comparing it with the standard form \(y^2=4px\), we get
$$4p=16$$
Which yields \(p=4\).
Therefore, the focus of the parabola is at \(F(4,0)\), and its directrix is \(x=-4\).
Find an equation of the parabola that has its vertex at the origin with the axis of symmetry lying on the y-axis, and passes through the point \(P(2,-5)\). What are the focus and directrix of the parabola?
Solution:
The equation of the given parabola is given by \(y=ax^2\). To calculate the value of \(a\) we use the condition that \(P(2,-5)\) lies on this parabola. This point will satisfy the equation, which gives \(-5=a(2)^2\) which implies that \(a=\dfrac{-5}{4}\).
Therefore, substituting the value of \(a\) in the equation of the parabola, we get
$$y=\dfrac{-5}{4}x^2$$
For finding the focus of the parabola, notice that it has the form \(F(0,p)\). The value of \(p\) can be calculated thanks to \(a\), as follows:
$$p=\dfrac{1}{4a}=\dfrac{-1}{5}$$
Therefore, the focus is \(F\left(0,-\dfrac{1}{5}\right)\). And the directrix will be \(y=-p\), which turns out to be \(y=\dfrac{1}{5}\).
Parabolas - Key takeaways
- A parabola is a set of points whose distance from a fixed line is equal to the distance from a fixed point.
- The point from which the distance is measured to define a parabola is known as the focus of the parabola.
- The fixed line from which the other distance is measured is known as the Directrix and is always perpendicular to the line segment formed by the vertex and the focus of the parabola.
- The standard parabola oriented with the x-axis is described by the equation \(y^2=4px\) and the one which is oriented with the y-axis is described by \(x^2=4py\).
- The equation of a parabola can be described by the set of parametric equations: \(x=pt^2\), \(y=2pt\) which basically gives us the form \(y^2=4px\).
- Parabola’s reflective property is used in radio telescopes, the headlights of automobiles, satellite dishes, etc.
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