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        riddles >> easy >> A Tangent to a Parabola
        
(Message started by: rloginunix on Oct 29^th, 2016, 8:01am)

Title: A Tangent to a Parabola
Post by rloginunix on Oct 29^th, 2016, 8:01am

A Tangent to a Parabola

In his beautiful work "Horologium Oscillatorium" (1673), Proposition 15, the Dutch mathematician and physicist Christiaan Huygens describes the mechanics of the straight edge and compass construction of a tangent to a cycloid. We are not as ambitious but, hopefully, as engaging:

given a graph of a parabola f(x) = ax² and an arbitrary point P(x₀, y₀) on that graph, describe a procedure of a straight edge and compass construction of a tangent http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/tau.gif to the graph of f(x) passing through P if (given and orthogonal) axes Ox and Oy have no gradation marks on them (no line segment of unit length):

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_ttoap.png

Extra for experts:

2) generalize

3) find the radius of a circle centered on Oy and tangent (internally) to the graph of f(x) at P

4) find the radius of the largest circle which, when rolled down one horn of the graph of f(x), will not get stuck near O but will rather roll right through it up and along the other horn

4') see if you can spot a connection between 4) and Off the Edge of the Earth (https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1413090975)

Title: Re: A Tangent to a Parabola
Post by towr on Oct 30^th, 2016, 11:24am

Maybe something like
[hide]
draw y=x
construct line parallel to y=2x through intersection
construct focal-point of parabola (reverse of following steps)
draw line parallel to x=0 through P
draw line from focal point through P
bisect angle between previous two lines
draw line perpendicular to previous line through P
[/hide]

Title: Re: A Tangent to a Parabola
Post by rloginunix on Oct 30^th, 2016, 4:36pm

I like it: a delicate step around the absence of a line segment of unit length plus, as a bonus, the location of [hide]parabola's focus[/hide].

towr's work is rendered here (https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1396709569;start=175#186) (the file name carries towr's credentials, the drawing is zoomed in a bit with P lowered with respect to the original, to avoid the clutter):

[hide]l1 = x[/hide], constructable, left as an exercise to the reader
[hide]l2 = 2x[/hide], constructable, ditto
[hide]l3[/hide]
[hide]F - parabola's focus[/hide], constructable, left as an exercise to the reader ([hide]double the angle DAC[/hide])
[hide]l4 and l5[/hide], in the never ending chase for an optimization, [hide]the last step may be shaved off: the bisector of the angle FPB is it[/hide].

Nice [hide]non-analysis[/hide] approach.

With one algorithm on the books, using [hide]an analysis-based argument, for an interpretation of a tangent,[/hide] the number of steps may be reduced.

Title: Re: A Tangent to a Parabola
Post by towr on Oct 30^th, 2016, 11:55pm

Well, I wouldn't say it's completely [hide]non-analytic[/hide], after all I had to get it from somewhere that [hide]the tangent where y=x crosses y=ax^2 is parallel to y=2x[/hide]

Reconsidering the morning after, I now realize you can cut out a significant number of steps [hide]because it's a matter of scaling one axis to make A=P[/hide]
So [hide]
construct point Q = (x₀, 2y₀)
construct line through P parallel to OQ.
[/hide]

Title: Re: A Tangent to a Parabola
Post by Grimbal on Oct 31^st, 2016, 3:43am

Can we use the axes? If so:
[hide]
- build P' on the other horn such that OP = OP',
- A = the intersection of the y axis and PP',
- B = reflection of A thru O (i.e. on the negative y axis with OA = OB)
- PB is the tangent.

It is like towr's solution except that the line (0,0)-(x₀,2y₀) is replaced by (0,-y₀)-(x₀,y₀).

[/hide]

Title: Re: A Tangent to a Parabola
Post by rloginunix on Oct 31^st, 2016, 8:28am

@Grimbal: yes, absolutely. Grimbal's solution (https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1396709569;start=175#187).

@towr: agreed. towr's second solution (https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1396709569;start=175#188).

By [hide]analysis-based[/hide] I meant the justification - and towr pretty much has it - [hide]instead of scaling along the Oy, scale along the Ox - from the right triangle[/hide] ...

Title: Re: A Tangent to a Parabola
Post by SWF on Nov 6^th, 2016, 1:46pm

A generalization would be for any power, n, and the curve a*xⁿ, draw the line through (0,-(n-1)*y₀), as in Grimbal's solution. You don't even need the curve for the construction, just the point, origin, and one of the axes.

Title: Re: A Tangent to a Parabola
Post by rloginunix on Nov 6^th, 2016, 3:19pm

Yes.

Correct.

By analytic justification I meant the textbookish observation that [hide]geometrically the first derivative is the slope (trigonometric tangent) of the tangent (straight line)[/hide]. Which means that in the right triangle PBX (first towr's drawing) where X is not shown - the point of intersection of the tangent and Ox [hide]BX = PB/tan(alpha) = ax²/2ax = x/2[/hide], etc.

Title: Re: A Tangent to a Parabola
Post by SWF on Nov 7^th, 2016, 5:32pm

One way to construct the focus is to make the line y=x/2. Where it cross the parabola has the y coordinate of the focus, y_F. This point is distance 2*y_F from the focus and same distance from the directrix, while the origin is distance y_F from both.

Title: Re: A Tangent to a Parabola
Post by rloginunix on Nov 8^th, 2016, 10:27am

Nice (https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1396709569;start=175#189).