# symmetry of parabola

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and

The idea that a parabolic reflector could produce an image was already well known before the invention of the reflecting telescope.

{\displaystyle f={\tfrac {c^{2}}{16d}}} 1 ,

If light travels along the line CE, it moves parallel to the axis of symmetry and strikes the convex side of the parabola at E. It is clear from the above diagram that this light will be reflected directly away from the focus, along an extension of the segment FE. The plane containing the circle If tangents to the parabola are drawn through the endpoints of any of these chords, the two tangents intersect on this same line parallel to the axis of symmetry (see Axis-direction of a parabola). {\displaystyle x=x_{2}}   16 {\displaystyle m_{1}-m_{2}.

=

These correspond to the explicit formula y = xp/q for a positive fractional power of x. ,

1

=

Q

Similarly, the structures of parabolic arches are purely in compression.

. {\displaystyle 4fd=\left({\tfrac {c}{2}}\right)^{2}}

y

{\displaystyle m_{0}}

{\displaystyle \sigma } {\displaystyle d}

1

0 O ) a

, the axis of symmetry is a vertical line

S is the focus, and V is the principal vertex of the parabola VG.

y 2

The right side of the diagram shows part of this parabola.

c )

with vertex c

Varsity Tutors connects learners with experts. y {\displaystyle \cos(3\alpha )=4\cos(\alpha )^{3}-3\cos(\alpha )} B is the midpoint of FC, so its y coordinate is zero, thus it lies on the x axis.

y

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0

Q , the angle between two lines of equations Conversely, two tangents that intersect on the directrix are perpendicular. ,

, t Draw perpendicular ST intersecting BQ, extended if necessary, at T. At B draw the perpendicular BJ, intersecting VX at J. →

4

0

y

d = , which has its vertex at the origin, opens upward, and has focal length f (see preceding sections of this article).

or, equivalently, such that P