Analysis puzzle
 

Here is a puzzle I found on some mathforum. I have not found a neat solution other than brute analysis (which, although some may consider this neat, is not really elegant).

Consider the equation y=x^2
For points (x,y) in the plane with y<x^2 we can draw two tangents to this curve, enclosing a certain area. Describe the set of points for which the area is constant.

No takers!
 

:smile:
Since no one has attempted your problem, kindly give us your solution, no matter how 'brutal' it may be.

I for one give up (though it can be worked out by calculus).

I have resurrected this thread so that others who may have missed it will consider it and give a solution.

However it will be interesting to know your way of solving it.

In the meantime I will consult my books on analysis as Im now out of touch with it.

Mally :coffee:

Without doing any work, my guess is y=x[sup]2[/sup]-c

David

[QUOTE=davieddy;102757]Without doing any work, my guess is y=x[sup]2[/sup]-c

David[/QUOTE]

Your guess is correct I reckon. I won't show the gory working, but I calculate the area in question (for a point (x,y), x^2 > y) to be

2/3(x[sup]2[/sup] - y)[sup](3/2)[/sup]

Chris

THX. So c = (3*area/2)[sup](2/3)[/sup]

Not going to show the gory details either. It is relatively simple to calculate the tangents to y=x^2 from a point (a,b) for which a>b^2, then integration and some triangle substraction will give an answer. But after finding the answer, which is just a translation of the y=x^2, I felt somewhat stupid and was certain that there was an easy geometrical explanation.

Have not found one yet so I just continue to feel silly,

Kees

Finally I found something. Well, to be more precise, Archimedes found something.
Take two points A, B on a parabola and let the tangents to the parabola at A and B intersect
at S. Then the area enclosed by the chord AB and the parabola is 2/3 of the area of the triangle ABS. The area which we are considering is the complentary 1/3 part of the triangle.
So the question can transformed to the question of finding the set of all such triangles with constant area.
Have not worked out the details yet, but this seems a good way to start.

The tangents from (a,0) hit y=x[sup]2[/sup] at (0,0) and
(2a,4a[sup]2[/sup]).
Shearing the diagram keeping the line x=a fixed maintains areas,
tangents and the shape of the parabola. It just shifts the apex.
I think the locus of the apex offers a promising approach.

David

PS Perhaps starting with the tangents from (0,-c) is even simpler.
(Shear keeping y axis fixed)

Yes, Since the sheared parabola passes through the origin,
the locus of its apex is obviously y=-x[sup]2[/sup]

I know the whole point of this thread is to avoid "gory" details,
but formally, the "shearing" transformation I am considering is:

x' = x
y' = y+kx
where k is a constant real number.

Are you paying attention Mally?

David

[QUOTE=davieddy;103038]I know the whole point of this thread is to avoid "gory" details,
but formally, the "shearing" transformation I am considering is:

x' = x
y' = y+kx
where k is a constant real number.

Are you paying attention Mally?

David[/QUOTE]

The problem is not put properly. The methods of attack are even worse.
To me its a waste of time.
"Shearing transformation"? Ha! ha ! Ha!
Mally :coffee: