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. 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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2 Derivation of the focus 3 Reflective property of the tangent 4 Constructing a parabola 5 External links |
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
Definitions and overview
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.
The parabola is an inverse transform of a cardioid.
Equations (Cartesian):
Vertical axis of symmetry:
Horizontal axis of symmetry:
Quadratic:
Equations (parametric):
See also: Ellipse, Hyperbola, Paraboloid.
Given a parabola parallel to the y-axis with vertex (0,0) and with equation
Let F denote the focus, and let Q denote the point at (x,-f). Line FP has the same length as line QP.
The tangent of the parabola described by equation (1) has slope
It follows that and 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 .
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 , so when it bounces off, its angle of inclination must be equal to . 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.)Derivation of the focus
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.
Square both sides,
Cancel out terms from both sides,
Cancel out the x2 from both sides (x is generally not zero),
Now let p=f and the equation for the parabola becomes
Q.E.DReflective property of the tangent
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:
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.