# Parabola

A parabola (from the Greek: παραβολή) is a conic section generated by the intersection of a cone and a plane tangent to the cone or parallel to some plane tangent to the cone. If the plane is itself tangent to the cone, one would obtain a degenerate parabola, a line. A parabola can also be defined as locus of points which are equidistant from a given point (the focus) and a given line (the directrix).

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## Definitions and overview

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Parabola_showing_focus_and_reflective_property.png
A graph showing the reflective property and the equidistant focus (blue) and directrix (green)

In Cartesian coordinates, a parabola with an axis parallel to the y axis with vertex (h, k), focus (h, k + p), and directrix y = k - p, with p being the distance from the vertex to the focus, has the equation

[itex](x - h)^2 = 4p(y - k) \,[itex]

or, alternatively

[itex](y - k) = \frac{1}{4p}(x-h)^2 \,[itex]

A parabola may also be characterized as a conic section with an eccentricity of 1. As a consequence of this, all parabolas are similar. A parabola can also be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction. The parabola is an inverse transform of a cardioid.

A parabola has a single axis of reflective symmetry, which passes through its focus and is perpendicular to its directrix. The point of intersection of this axis and the parabola is called the vertex. A parabola spun about this axis in three dimensions traces out a shape known as a paraboloid of revolution.

The parabola is found in numerous situations in the physical world (see below).

### Equations

#### Cartesian

• Vertical axis of symmetry:
[itex](x - h)^2 = 4p(y - k) \quad [itex]
• Horizontal axis of symmetry:
[itex](y - h)^2 = 4p(x - k) \quad [itex]
• Quadratic (vertical axis of symmetry):
[itex]y = ax^2 + bx + c \,[itex]
[itex]\mbox{with }a = \frac{1}{4p}; \ \ b = \frac{-h}{2p}; \ \ c = \frac{h^2}{4p} + k[itex] and the vertex [itex]\left( \frac{-b}{2a},\ \frac{4ac - b^2}{4a} \right)[itex].
• Quadratic (horizontal axis of symmetry):
[itex]x = ay^2 + by + c \;[itex]
a, b, and c are the same as above. The coordinates of the vertex are reversed.

#### Parametric

[itex]x = 2pt + h \,[itex]
[itex]y = pt^2 + k \,[itex]

#### Semi-latus rectum and polar coordinates

In polar coordinates, a parabola with the focus at the origin and the top on the negative x-axis, is given by the equation

[itex]r (1 - \cos \theta) = l \,[itex]

where l is the semi-latus rectum: the distance from the focus to the parabola itself, measured along a line perpendicular to the axis. Note that this is twice the distance to the top.

#### Gauss-mapped form

A Gauss-mapped form: [itex](\tan^2\phi,2\tan\phi)[itex] has normal [itex](\cos\phi,\sin\phi)[itex].

## Derivation of the focus

Given a parabola parallel to the y-axis with vertex (0,0) and with equation

then there is a point (0,f) — the focus — such that any point P on the parabola will be equidistant from both the focus and a line perpendicular to the axis of symmetry of the parabola (the linea directrix), in this case parallel to the x axis. Since the vertex is one of the possible points P, it follows that the linea directrix passes through the point (0,-f). So for any point P=(x,y), it will be equidistant from (0,f) and (x,-f). It is desired to find the value of f which has this property.

Let F denote the focus, and let Q denote the point at (x,-f). Line FP has the same length as line QP.

[itex] \| FP \| = \sqrt{ x^2 + (y - f)^2 }, [itex]
[itex] \| QP \| = y + f. [itex]
[itex] \| FP \| = \| QP \| [itex]
[itex] \sqrt{x^2 + (a x^2 - f)^2 } = a x^2 + f \qquad [itex]

Square both sides,

[itex] x^2 + (a x^2 - f)^2 = (a x^2 + f)^2 \qquad [itex]
[itex] = a^2 x^4 + f^2 + 2 a x^2 f \quad [itex]
[itex] x^2 + a^2 x^4 + f^2 - 2 a x^2 f = a^2 x^4 + f^2 + 2 a x^2 f \quad [itex]

Cancel out terms from both sides,

[itex] x^2 - 2 a x^2 f = 2 a x^2 f, \quad [itex]
[itex] x^2 = 4 a x^2 f. \quad [itex]

Cancel out the x2 from both sides (x is generally not zero),

[itex] 1 = 4 a f \quad [itex]
[itex] f = {1 \over 4 a } [itex]

Now let p=f and the equation for the parabola becomes

[itex] x^2 = 4 p y \quad [itex]

## Reflective property of the tangent

The tangent of the parabola described by equation (1) has slope

[itex] {dy \over dx} = 2 a x = {2 y \over x} [itex]

This line intersects the y-axis at the point (0,-y) = (0, - a x2), and the x-axis at the point (x/2,0). Let this point be called G. Point G is also the midpoint of points F and Q:

[itex] F = (0,f), \quad [itex]
[itex] Q = (x,-f), \quad [itex]
[itex] {F + Q \over 2} = {(0,f) + (x,-f) \over 2} = {(x,0) \over 2} = ({x \over 2}, 0). [itex]

Since G is the midpoint of line FQ, this means that

[itex] \| FG \| \cong \| GQ \|, [itex]

and it is already known that P is equidistant from both F and Q:

[itex] \| PF \| \cong \| PQ \|, [itex]

and, thirdly, line GP is equal to itself, therefore:

[itex]\Delta FGP \cong \Delta QGP[itex]

It follows that [itex] \angle FPG \cong \angle GPQ [itex].

Line QP can be extended beyond P to some point T, and line GP can be extended beyond P to some point R. Then [itex] \angle RPT [itex] and [itex] \angle GPQ [itex] are vertical, so they are equal (congruent). But [itex] \angle GPQ [itex] is equal to [itex] \angle FPG [itex]. Therefore [itex] \angle RPT [itex] is equal to [itex] \angle FPG [itex].

The line RG is tangent to the parabola at P, so any light beam bouncing off point P will behave as if line RG were a mirror and it were bouncing off that mirror.

Let a light beam travel down the vertical line TP and bounce off from P. The beam's angle of inclination from the mirror is [itex] \angle RPT [itex], so when it bounces off, its angle of inclination must be equal to [itex] \angle RPT [itex]. But [itex] \angle FPG [itex] has been shown to be equal to [itex] \angle RPT [itex]. Therefore the beam bounces off along the line FP: directly towards the focus.

Conclusion: Any light beam moving vertically downwards in the concavity of the parabola (parallel to the axis of symmetry) will bounce off the parabola moving directly towards the focus. (See parabolic reflector.)

## Parabolae in the physical world

In nature, approximations of parabolae and paraboloids are found in many diverse situations. The most well-known instance of the parabola in the history of physics is the trajectory of particle or body in motion under the influence of a uniform gravitational field without air resistance (for instance, a baseball flying through the air, neglecting air friction). The parabolic trajectory of projectiles was discovered experimentally by Galileo in the early 17th century, who performed experiments with balls rolling on inclined planes. The parabolic shape for projectiles was later proven mathematically by Isaac Newton. For objects extended in space, such as a diver jumping from a diving board, the object itself follows a complex motion as it rotates, but the center of mass of the object nevertheless forms a parabola. As in all cases in the physical world, the trajectory is always an approximation of a parabola. The presence of air resistance always distorts the shape, for example, although at low speeds, the shape is a good approximation of a parabola. At higher speeds, such as in ballistics, the shape is highly distorted and does not resemble a parabola.

Missing image
DSCN8987_orangeparabola_e.jpg
Parabolic shape formed by the surface of a liquid under rotation

Another situation in which parabola may arise in nature is in two-body orbits, for example, of a small planetoid or other object under the influence of the gravitation of the sun. In such a case, parabolic orbits are a special case that are in nature. Orbits that form a hyperbola or an ellipse are much more common. In fact, the parabolic orbit is the borderline case between those two types of orbit.

Approximations of parabolas are also found in the shape of cables of suspension bridges. Freely hanging cables do not describe parabolas, but rather catenary curves. Under the influence of a uniform load (for example, the deck of bridge), however, the cable is deformed towards a parabola.

Paraboloids arise in several physical situations as well. The most well-known instance is the parabolic reflector, which is a mirror or similar reflective device that concentrates light or other forms of electromagnetic radiation to a common focal point. The principle of the parabolic reflector was discovered by the geometer Archimedes in the 3rd century B.C., who constructed parabolic mirrors to defend Syracuse against the Roman fleet, by concentrating the sun's rays to set fire to the decks of the Roman ships. The principle was applied to telescopes in the 17th century. Today, paraboloid reflectors can be commonly observed throughout much of the world in microwave and satellite dish antennas.

Paraboloid are also observed in the surface of a liquid confined to a container and rotated around the central axis. In this case, the centrifugal force (a fictitious force), causes the surface of the liquid to climb the walls of the container, forming a parabolic surface.

## Constructing a parabola

A parabola can be constructed geometrically as follows: draw focus F, vertex, linea directrix q, and linea verticis r (through the vertex, parallel to linea directrix). Choose a point Q1 on linea directrix. Draw line FQ1 which intersects linea verticis at R1. A line (through R1 and perpendicular to FQ1 ) will intersect another line (through Q1 and perpendicular to linea directrix) at point P1. Point P1 is on the parabola, and line R1P1 is tangential to the parabola. Choose another point Q2 on linea directrix and repeat the steps of the paradigm above to obtain P2. Continue with points [itex] Q_3, P_3, Q_4, P_4, Q_5, P_5 [itex], et cetera. If the points [itex] Q_1, Q_2, Q_3, ... [itex] were drawn in a sequence, then the points [itex] P_1, P_2, P_3 ... [itex] can be connected sequentially to draw the parabola.

### By paper folding

Draw a straight line (the directrix) on a piece of paper, and a point (the focus) somewhere not on the line. Then fold the paper over so that the focus point touches the directrix line and crease the fold. Also crease the paper vertically where the focus point touches the directrix line. The point where these two lines intersect is a point on the parabola. Do this several timesto get more points on the parabola. The envelope formed by the creases will make a nice parabola.

You can make an ellipse or hyperbola similarly by using a circle and a point.

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