Parabola

For other uses, see Parabola (disambiguation).

A parabola

The parabola (from the Greek: παραβολή) is a conic section generated by the intersection of a right circular conical surface and a plane parallel to a generating straight line of that surface. A parabola can also be defined as locus of points in a plane which are equidistant from a given point (the focus) and a given line (the directrix).

A particular case arises when the plane is tangent to the conical surface. In this case, the intersection is a degenerate parabola consisting of a straight line.

Definitions and overview

A graph showing the reflective property, the directrix (green), and the lines connecting the focus and directrix to the parabola (blue)

Analytic geometry equations

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

$(x-h)^{2}=4p(y-k)\,$

or, alternatively

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

More generally, a parabola is a curve in the Cartesian plane defined by an irreducible equation of the form

$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\,$

such that $B^{2}=4AC\,$, where all of the coefficients are real, where A and/or C is non-zero, and where more than one solution, defining a pair of points (x, y) on the parabola, exists. That the equation is irreducible means it does not factor as a product of two not necessarily distinct linear factors.

Other geometric definitions

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. In this sense, a parabola may be considered an ellipse that has one focus at infinity. 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

(with vertex (h, k) and distance p between vertex and focus - note that if the vertex is below the focus, or equivalently above the directrix, p is positive, otherwise p is negative; similarly with horizontal axis of symmetry p is positive if vertex is to the left of the focus, or equivalently to the right of the directrix)

Cartesian

Vertical axis of symmetry

$(x-h)^{2}=4p(y-k)\,$

$y=a(x-h)^{2}+k\,$

$y=ax^{2}+bx+c\,$

${\mbox{where }}a={\frac {1}{4p}};\ \ b={\frac {-h}{2p}};\ \ c={\frac {h^{2}}{4p}}+k;\ \$

$h={\frac {-b}{2a}};\ \ k={\frac {4ac-b^{2}}{4a}}$.

$x(t)=2pt+h;\ \ y(t)=pt^{2}+k\,$

Horizontal axis of symmetry

$(y-k)^{2}=4p(x-h)\,$

$x=a(y-k)^{2}+h\,$

$x=ay^{2}+by+c\,$

${\mbox{where }}a={\frac {1}{4p}};\ \ b={\frac {-k}{2p}};\ \ c={\frac {k^{2}}{4p}}+h;\ \$

$h={\frac {4ac-b^{2}}{4a}};\ \ k={\frac {-b}{2a}}$.

$x(t)=pt^{2}+h;\ \ y(t)=2pt+k\,$

Bülent Bircan Fan Club

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

$r(1-\cos \theta )=l\,$

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 from the focus to the apex of the parabola or the perpendicular distance from the focus to the latus rectum.

Gauss-mapped form

A Gauss-mapped form: $(\tan ^{2}\phi ,2\tan \phi )$ has normal $(\cos \phi ,\sin \phi )$.

Derivation of the focus

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

$y=ax^{2},\qquad \qquad \qquad (1)$

Parabolic Curve Showing Directrix. Its Equation: $x^{2}=8y$

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.

$\|FP\|={\sqrt {x^{2}+(y-f)^{2}}},$

$\|QP\|=y+f.$

$\|FP\|=\|QP\|$

${\sqrt {x^{2}+(ax^{2}-f)^{2}}}=ax^{2}+f\qquad$

Square both sides,

$x^{2}+(ax^{2}-f)^{2}=(ax^{2}+f)^{2}\qquad$

$=a^{2}x^{4}+f^{2}+2ax^{2}f\quad$

$x^{2}+a^{2}x^{4}+f^{2}-2ax^{2}f=a^{2}x^{4}+f^{2}+2ax^{2}f\quad$

Cancel out terms from both sides,

$x^{2}-2ax^{2}f=2ax^{2}f,\quad$

$x^{2}=4ax^{2}f.\quad$

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

$1=4af\quad$

$f={1 \over 4a}$

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

$x^{2}=4py\quad$

Q.E.D.

Reflective property of the tangent

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

${dy \over dx}=2ax={2y \over x}$

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:

$F=(0,f),\quad$

$Q=(x,-f),\quad$

${F+Q \over 2}={(0,f)+(x,-f) \over 2}={(x,0) \over 2}=({x \over 2},0).$

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

$\|FG\|\cong \|GQ\|,$

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

$\|PF\|\cong \|PQ\|,$

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

$\Delta FGP\cong \Delta QGP$

It follows that $\angle FPG\cong \angle GPQ$.

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

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 $\angle RPT$, so when it bounces off, its angle of inclination must be equal to $\angle RPT$. But $\angle FPG$ has been shown to be equal to $\angle RPT$. 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.)

Analyzing the Parabola

Part I: Introduction

Analyzing the parabola is a way to solve quadratic equations that depends on the symmetry and other characteristics of the parabola. It is easiest to use this method when you are given a problem in which you must solve for the vertex first. This method can also be exploited to derive a second quadratic formula.

Part II: The “Height” of the Parabola

This is one of the most crucial concepts of using this method. The “height” of a parabola is really the absolute value of the distance along the axis of symmetry between the vertex of the function and the line connecting two solutions of the quadratic equation for any value of x. For example (y=2x²) the height between the origin (the vertex in this case) when x=2 is 8. In this paper, the formula for the height will be given as $h=-ax^{2}$.

Part III: Examples of Applying the Method

Example 1: 2x²+3x-4

• Step 1: Find the vertex

vtx = (-0.75, -5.125)

• Step 2: Set the height of the parabola equal to the y-vertex value and solve for x

h = - ax²

-5.125 = -2x²

x = ±√(-5.125/-2) ≈ ±1.601

• Step 3: Up until now the x value of the vertex has been presumed to be zero. We must now add the x value in the vertex to the solutions so that they are relative to the x-axis, not the line of symmetry.

x ≈ ±1.601 - 0.75 = 0.851, -2.351

Example 2: 9x²+2x+4

• Step 1: Find the vertex

vtx = (-0.1111, 3.8888)

• Step 2: Set the height of the parabola equal to the y-vertex value and solve for x

h = -ax²

3.8888 = -9x²

x = ±√(3.8888/-9) ≈ ±0.6573i

• Step 3: Up until now the x value of the vertex has been presumed to be zero. We must now add the x value in the vertex to the solutions so that they are relative to the x-axis, not the line of symmetry.

x ≈ ±0.6573i-0.1111 = -0.1111-0.6573i, -0.1111+0.6573i

Part IV: The Second Quadratic Formula

Following these steps, it is possible to derive a second quadratic formula.

• Step 1: Find the vertex

vtx = (j, k)

• Step 2: Set the height of the parabola equal to the y-vertex value and solve for x

h = -ax²

k = -ax²

x = ±√(k/-a)

• Step 3: Up until now the x value of the vertex has been presumed to be zero. We must now add the x value in the vertex to the solutions so that they are relative to the x-axis, not the parabola’s line of symmetry.

x = j±√(k/-a)

The a coefficient must be of the opposite sign in this formula or else the roots to an equation with two real solutions will be imaginary.

Part V: Sample Problems

• Example 1

vtx = (-3, -9)

a = 2

x = j±√(k/-a)

x = -3 ± √(-9/-2) ≈ -0.8787, -5.1213

• Example 2

vtx = (-2.25, 8.125)

a = -2

x = j±√(k/-a)

x = -2.25 ± √(8.125/2) ≈ 0.2344, 4.26556

Parabolas in the physical world

In nature, approximations of parabolas 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 a 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, for example, always distorts the shape 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.

Parabolic shape formed by the surface of a Newtonian 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. Such parabolic orbits are a special case that are rarely found 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.

Parabolic arches in Antoni Gaudí's Casa Milà.

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 in the 3rd century BC by the geometer Archimedes, who, according to a legend of debatable veracity, 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.

Paraboloids 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 causes the liquid to climb the walls of the container, forming a parabolic surface. This is the principle behind the liquid mirror telescope.

See also

• Conic section
• Catenary
• Paraboloid
• Hyperbola
• Ellipse

External links

• Weisstein, Eric W. "Parabola". MathWorld.
• Archimedes Triangle and Squaring of Parabola at cut-the-knot
• Two Tangents to Parabola at cut-the-knot
• Parabola As Envelope of Straight Lines at cut-the-knot
• Parabolic Mirror at cut-the-knot
• Three Parabola Tangents at cut-the-knot
• Two Tangents to Parabola at cut-the-knot
• Focal Properties of Parabola at cut-the-knot
• Parabola As Envelope II at cut-the-knot
• Parabola Construction - An interactive sketch showing how to trace a parabola. (Requires Java.)
• Quadratic Bezier Construction - An interactive sketch showing how to trace the quadratic Bezier curve (a parabolic segment). (Requires Java.)
• More Interactive Parabola Construction (Java-enabled)