Parabola
A parabola 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. In other words, a parabola is the locus of points which are equidistant from a given point (the focus) and a given line (the directrix).
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 has the equation
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.
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. See also parabolic reflector.
A particle or body in motion under the influence of a uniform gravitational field (for instance, a baseball flying through the air, neglecting air friction) follows a parabolic trajectory.
Vertical axis of symmetry:
Horizontal axis of symmetry:
Quadratic:
Equations (Parametric):

See also: Ellipse,
Hyperbola,
Paraboloid.
Given a parabola parallel to the yaxis 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.




Square both sides,



Cancel out terms from both sides,


Cancel out the x^{2} from both sides (x is generally not zero),


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

Quod Erat Demonstrandum.
The tangent of the parabola described by equation (1) has slope

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



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

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

and, thirdly, line GP is equal to itself, therefore the triangles ΔFGP and ΔQGP are congruent.
If follows that the angles FPG and GPQ are equal. Line QP can be extended beyond P to some point T, and line GP can be extended beyond P to some point R. Then and are vertical, so they are equal (congruent). But is equal to . Therefore is equal to angle FPQ.
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 RPT, so when it bounces off, its angle of inclination must be equal to RPT. But has been shown to be equal to . 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.)
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 Q_{1} on linea directrix. Draw line FQ_{1} which intersects linea verticis at R_{1}. A line (through R_{1} and perpendicular to FQ_{1} ) will intersect another line (through Q_{1} and perpendicular to linea directrix) at point P_{1}. Point P_{1} is on the parabola, and line R_{1}P_{1} is tangential to the parabola. Choose another point Q_{2} on linea directrix and repeat the steps of the paradigm above to obtain P_{2}. Continue with points , et cetera. If the points were drawn in a sequence, then the points can be connected sequentially to draw the parabola.
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