Parabola

A parabola (from the Greek language: ''παραβολή'') is a conic section generated by the intersection of a Cone (geometry) and a plane (mathematics) 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 mathematical degeneracy parabola, a line_(mathematics). A parabola can also be defined as locus (mathematics) of points which are equality (mathematics)distance from a given point (the Focus (geometry)) and a given line (the directrix). ==Definitions and overview== 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 :$(x\; -\; h)^2\; =\; 4p(y\; -\; k)\; \backslash ,$ or, alternatively :$(y\; -\; k)\; =\; \backslash frac\{1\}\{4p\}(x-h)^2\; \backslash ,$ A parabola may also be characterized as a conic section with an eccentricity (mathematics) of 1. As a consequence of this, all parabolas are similar. A parabola can also be obtained as the limit of a sequence of ellipse where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction. The parabola is an inversive geometry 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: :$(x\; -\; h)^2\; =\; 4p(y\; -\; k)\; \backslash quad$ *Horizontal axis of symmetry: :$(y\; -\; h)^2\; =\; 4p(x\; -\; k)\; \backslash quad$ *Quadratic (vertical axis of symmetry): :$y\; =\; ax^2\; +\; bx\; +\; c\; \backslash ,$ ::$\backslash mbox\{with\; \}a\; =\; \backslash frac\{1\}\{4p\};\; \backslash \; \backslash \; b\; =\; \backslash frac\{-h\}\{2p\};\; \backslash \; \backslash \; c\; =\; \backslash frac\{h^2\}\{4p\}\; +\; k$ and the vertex $\backslash left(\; \backslash frac\{-b\}\{2a\},\backslash \; \backslash frac\{4ac\; -\; b^2\}\{4a\}\; \backslash right)$. *Quadratic (horizontal axis of symmetry): :$x\; =\; ay^2\; +\; by\; +\; c\; \backslash ;$ :''a'', ''b'', and ''c'' are the same as above. The coordinates of the vertex are reversed. ====Parametric==== :$x\; =\; 2pt\; +\; h\; \backslash ,$ :$y\; =\; pt^2\; +\; k\; \backslash ,$ ====Semi-latus rectum and polar coordinates==== In coordinates (elementary mathematics), a parabola with the focus at the origin and the top on the negative ''x''-axis, is given by the equation : $r\; (1\; -\; \backslash cos\; \backslash theta)\; =\; l\; \backslash ,$ 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 map form: $(\backslash tan^2\backslash phi,2\backslash tan\backslash phi)$ has normal $(\backslash cos\backslash phi,\backslash sin\backslash phi)$. ==See also== *Paraboloid *Ellipse *Hyperbola *conic section ==Derivation of the focus== Given a parabola parallel to the ''y''-axis with vertex (0,0) and with equation :$y\; =\; a\; x^2,\; \backslash qquad\; \backslash qquad\; \backslash qquad\; (1)$ 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''. :$\backslash |\; FP\; \backslash |\; =\; \backslash sqrt\{\; x^2\; +\; (y\; -\; f)^2\; \},$ :$\backslash |\; QP\; \backslash |\; =\; y\; +\; f.$ :$\backslash |\; FP\; \backslash |\; =\; \backslash |\; QP\; \backslash |$ :$\backslash sqrt\{x^2\; +\; (a\; x^2\; -\; f)^2\; \}\; =\; a\; x^2\; +\; f\; \backslash qquad$ Square both sides, :$x^2\; +\; (a\; x^2\; -\; f)^2\; =\; (a\; x^2\; +\; f)^2\; \backslash qquad$ :::$=\; a^2\; x^4\; +\; f^2\; +\; 2\; a\; x^2\; f\; \backslash quad$ :$x^2\; +\; a^2\; x^4\; +\; f^2\; -\; 2\; a\; x^2\; f\; =\; a^2\; x^4\; +\; f^2\; +\; 2\; a\; x^2\; f\; \backslash quad$ Cancel out terms from both sides, :$x^2\; -\; 2\; a\; x^2\; f\; =\; 2\; a\; x^2\; f,\; \backslash quad$ :$x^2\; =\; 4\; a\; x^2\; f.\; \backslash quad$ Cancel out the ''x²'' from both sides (''x'' is generally not zero), :$1\; =\; 4\; a\; f\; \backslash quad$ :$f\; =\; \{1\; \backslash over\; 4\; a\; \}$ Now let ''p=f'' and the equation for the parabola becomes :$x^2\; =\; 4\; p\; y\; \backslash quad$ Q.E.D. ==Reflective property of the tangent== The tangent of the parabola described by equation (1) has slope :$\{dy\; \backslash over\; dx\}\; =\; 2\; a\; x\; =\; \{2\; y\; \backslash over\; x\}$ 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'': :$F\; =\; (0,f),\; \backslash quad$ :$Q\; =\; (x,-f),\; \backslash quad$ :$\{F\; +\; Q\; \backslash over\; 2\}\; =\; \{(0,f)\; +\; (x,-f)\; \backslash over\; 2\}\; =\; \{(x,0)\; \backslash over\; 2\}\; =\; (\{x\; \backslash over\; 2\},\; 0).$ Since ''G'' is the midpoint of line ''FQ'', this means that :$\backslash |\; FG\; \backslash |\; \backslash cong\; \backslash |\; GQ\; \backslash |,$ and it is already known that ''P'' is equidistant from both ''F'' and ''Q'': :$\backslash |\; PF\; \backslash |\; \backslash cong\; \backslash |\; PQ\; \backslash |,$ and, thirdly, line ''GP'' is equal to itself, therefore: :$\backslash Delta\; FGP\; \backslash cong\; \backslash Delta\; QGP$ It follows that $\backslash angle\; FPG\; \backslash cong\; \backslash angle\; GPQ$. Line ''QP'' can be extended beyond ''P'' to some point ''T'', and line ''GP'' can be extended beyond ''P'' to some point ''R''. Then $\backslash angle\; RPT$ and $\backslash angle\; GPQ$ are vertical, so they are equal (congruent). But $\backslash angle\; GPQ$ is equal to $\backslash angle\; FPG$. Therefore $\backslash angle\; RPT$ is equal to $\backslash angle\; FPG$. 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 $\backslash angle\; RPT$, so when it bounces off, its angle of inclination must be equal to $\backslash angle\; RPT$. But $\backslash angle\; FPG$ has been shown to be equal to $\backslash angle\; RPT$. 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.) ==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. 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, Italy against the Roman Empire 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 $Q\_3,\; P\_3,\; Q\_4,\; P\_4,\; Q\_5,\; P\_5$, et cetera. If the points $Q\_1,\; Q\_2,\; Q\_3,\; ...$ were drawn in a sequence, then the points $P\_1,\; P\_2,\; P\_3\; ...$ 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. ==See also== *Catenary ==External links== * [http://mathworld.wolfram.com/Parabola.html MathWorld: Parabola] * [http://www.cut-the-knot.org/Curriculum/Geometry/ArchimedesTriangle.shtml Archimedes Triangle and Squaring of Parabola] * [http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaLambert.shtml Two Tangents to Parabola] * [http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaEnvelope.shtml Parabola As Envelope of Straight Lines] * [http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaMirror.shtml Parabolic Mirror] * [http://www.cut-the-knot.org/Curriculum/Geometry/ThreeParabolaTangents.shtml Three Parabola Tangents] * [http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaLambert.shtml Two Tangents to Parabola] * [http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaFocal.shtml Focal Properties of Parabola] * [http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaMesh.shtml Parabola As Envelope II] Conic sections

Parabola

So, what does a parabola look like? What does it have to do with a parabolic mirror? How is it different in shape from a hyperbola or one end of an eccentric ellipse? :Here is a parabola: U :Here is another one: C :User:Dietary Fiber :: Hmm.. maybe a superimposed pic would be good. A parabola doesn't have asymptotes, a hyperbola does. An ellipse is curvy enough to close up again at the other side -- so it's a matter of curviness really. -- User:Tarquin 21:16 Mar 28, 2003 (UTC) I know all that, I was very good in math. I just want someone who's good at ''drawing'' to draw a picture of a parabola! --User:Ed Poor I am! U User:Dietary Fiber ---- Q: What is the origin of the name 'parabola'? Is there something to do with 'parallel'? In Japanese, parabola is called 放物線(Ho-Butsu-Sen), which means the curve(Sen) of thrown(Ho) object(Butsu). --User:HarpyHumming 20:54, 26 Feb 2004 (UTC) :It comes from the Greek words "para" (across) and "ballein" (to throw), so it's similar to the Japanese word. (Parabola is also the ancestor of "parable," the French word "parler," and its relative "parliament".) User:Adam Bishop 20:58, 26 Feb 2004 (UTC) ---- Why hasn't a simple y=x^2 been mentioned? I agree. There should be a section discussing how/why y=x^2 forms a parabola. User:Mpiff 03:51, 9 Dec 2004 (UTC) ---- User: Nobody_EDN 2004.10.22 Withdrawn because of lack of interst. Why aren't there more ways to produce a parabola than folding paper given??? The pencil and string method seems a good one to add. ---- ''By paper folding'' ''Draw a straight line on a piece of paper, and a point somewhere not on the line. Then fold the paper over so that the point touches the line and crease the fold. Do this several times. 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.'' These directions are hard to follow.

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