ParabolasFocus

A fixed point F (0 ,f) and the straight liney = -fhave been given.

P is any point and V is its projection at that line.

P(x , y) is chosen so, that FP = PV. It can be done in infinite ways.

What is the shape of the collections of points P, so what is the locus of P?

FP = √{(x-0)^{2}+ (y-f)^{2}} and PV =y+f

Sox^{2}+ (y-f)^{2}= (y + f)^{2}ory = x^{2}/(4f)

It is a parabolawith its minimum in (0 , 0).

In the math lesson is told, that such a parabola is presented byy = ax^{2}.

Obviouslya= 1/(4f) orf= 1/(4a) or 4af= 1.

The given line is called the directrix of the parabola.

F is called the focus andfthe focal length being the distance from F to the top.

Any shift of the parabola leads to the equationy = ax^{2}+bx + cor

A parallel beam of light falling on a parabolic mirror is reflected to ay = x^{2}/(4f) +bx + c

## Parabolic mirror

point F’on a line perpendicular to the axis of the parabola.

The light beam through F is reflected parallel to the main axis.

This way you find F’.

The parabola## De parabola star

y= (1/8)x^{2}+ 2 meets the points (-4 , 4) and (4 , 4).

The top is at (0 , 2) and F at (0 , 4 ), becausef= 2.

The tangent at (4 , 4) is the liney = x. The tangent at (-4 , 4) is the liney = -x.

Please, check it yourself.

Design the mirror image of this parabola relative to the x-axis.

Design the mirror image of this parabola relative toy = x.

Design the mirror image of this parabola relative toy= -x.

Then a ‘parabola star’ arises.

The adjoining parabolas touch each other at the vertices.

## If

Similarityais tall and sofis small, then the parabola is narrow and like a bullet.

Ifais small and sofis tall, then the parabola wide as an umbrella.

But still: all parabolas are similar.

Check the top of the 'pointy' parabola with a magnifying glass.The

parabola vergelijking f1 3 x^{2}1/12 2 (1/5) x^{2}5/4 fof parabola 2 is 15 × so tall as thefof parabola 1.

The (linear) magnification factor is 15.

All lengths in parabola 2 are 15 × the corresponding lengths in parabola 1.

## Consider the ↑ parabola

Parabolic wavey= -px(x-2a) fromx=0 up to and includingx=2a.

The maximum is (a , pa^{2}).

Find a new parabola by rotating this ↑ parabola 180° around point (2a, 0).

Then you get a congruent ↓ parabola which reaches fromx=2aup to and includingx=4a.

The minimum is ( 3a, -ap^{2}).

A complete 'wavelength' with length 4aarises.

This figure may be extended with as much congruent wavelength as wanted,

to the left and to the right.

The wave does not show any gap or jump, in particular at the points of intersection with thex-axis.

We ourselves made the wave continuous.

It can be shown by differential calculus, that the wave has not a kink atx= 2a

nor at any point 4afurther along the x-axis.The derivative ofLikewise, it can be understood that the steepness of the ↓ parabola aty= -px(x-2a) = -px^{2}+ 2apxisy’= -2px+ 2ap.

Ifx= 2a, theny’ = -2p.2a+ 2ap= -2ap.

So the slope of the ↑ parabola atx=2ais -2ap.x=2aequals -2apas well.

It means that the two parabolas have a common tangent.

This line is an inflection tangent to the wave.

The wave is differentiable at any point, especially atx= 0 + k.2a.

The parabola wave does not shows any kink.

If one would draw a sine wave with the same wavelength 4aand the same amplitudopa^{2}

in the same coordinate system, then a difference relative to the parabolic wave is hardly seen.

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