1. ## Find This Curve!

Consider the point that is the circle's maximum (so the point (0, radius), putting the origin at the circle's centre).

Find the curve traced by this point as the ellipse (the circle) becomes infinite on the vertical plane (infinitely tall, while remaining the same width).

In other words, if the circle is x^2/A^2 + y^2 = B^2, find the above described curve traced as A goes to infinity.

This is hard to describe, I wish I could draw a picture for you!

2. Why isn't it just the ray x = 0 for y >= "radius"?

Perhaps a better description would be good.

3. Hopefully you will be able to view the attachment.

The problem is to find the curve traced by the red dot as the height of the ellipse is allowed to go to infinity.

The distance along the curve from the blue dot to the red dot remains constant.

4. Originally Posted by Luc
Hopefully you will be able to view the attachment.

The problem is to find the curve traced by the red dot as the height of the ellipse is allowed to go to infinity.

The distance along the curve from the blue dot to the red dot remains constant.
The locus of all points which have a constant distance to a fixed is a circle line.

5. Originally Posted by earboth
The locus of all points which have a constant distance to a fixed is a circle line.
It's a constant distance along each curve to the blue point. Not a constant distance from the curve to the blue point ..... I had the same initial misunderstanding .....

6. Oops, sorry, should have clarified that a bit more.

If the initial circle's circumference is C, then the length of the segment from the blue dot to the red dot is C/4, and is constant (because we are considering the path of the red dot).

So I've had some ideas, but no solutions yet.

x^2/A^2 + y^2 = ((2L)/pi)^2

where L is the length along the curve from blue dot to red dot.

Since we're letting A go to infinity, we are dealing with x, y, and A as variables.
So my idea is to find x(A), y(A), then jiggle them around to get y(x).

We take L and equate it to the the integral giving the length of the curve, integrating from 0 to y(A),
so

L = [integral from 0 to y(A) of] dy x sqrt(1 + (dx/dy)^2)

then use the first relation to find the expressions for the derivatives and to get everything in terms of 2 variables...

Then you come out with

L = [integral from 0 to y(A) of] dy x sqrt(1 + (Ay)^2/(((2L)/pi)^2 - y^2))

...and that's as far as I'm willing to go for today.

PS if anyone could give me tips on how to post real math text like square roots, exponents, symbols and whatnot that would be great, since I dislike typing out all that garbage as much as you probably dislike deciphering it!

7. Originally Posted by Luc
PS if anyone could give me tips on how to post real math text like square roots, exponents, symbols and whatnot that would be great, since I dislike typing out all that garbage as much as you probably dislike deciphering it!
It is called la Tex here is a web link for you to look at.

Help Displaying a formula - Wikipedia, the free encyclopedia

8. Originally Posted by Luc
Oops, sorry, should have clarified that a bit more.

If the initial circle's circumference is C, then the length of the segment from the blue dot to the red dot is C/4, and is constant (because we are considering the path of the red dot).

So I've had some ideas, but no solutions yet.

x^2/A^2 + y^2 = ((2L)/pi)^2

where L is the length along the curve from blue dot to red dot.

Since we're letting A go to infinity, we are dealing with x, y, and A as variables.
So my idea is to find x(A), y(A), then jiggle them around to get y(x).

We take L and equate it to the the integral giving the length of the curve, integrating from 0 to y(A),
so

L = [integral from 0 to y(A) of] dy x sqrt(1 + (dx/dy)^2)

then use the first relation to find the expressions for the derivatives and to get everything in terms of 2 variables...

Then you come out with

L = [integral from 0 to y(A) of] dy x sqrt(1 + (Ay)^2/(((2L)/pi)^2 - y^2))

...and that's as far as I'm willing to go for today.

PS if anyone could give me tips on how to post real math text like square roots, exponents, symbols and whatnot that would be great, since I dislike typing out all that garbage as much as you probably dislike deciphering it!
Any integral set up to calculate the arc length along an ellipse is doomed not to have an exact answer in terms of a finite number of elementary functions. You might find this reference interesting:

Elliptic Arcs

This is of course a major blow to finding an exact equation for the locus by algebraic means. There are other problems too .....

9. OK so I threw calculus out the window and just solved it with geometry after I realized that the curve I was trying to find was elliptical (thanks to the article that mr.fantastic posted).
It was alot easier.

I'll just walk you through what I did, using the attached picture.

I called the constant distance along the curve from the previous image "L", and made (0,L) the vertice of the ellipse who's identity we are finding.
I kept the circle that I started with in the other picture and made it of radius "r", centred at (r,0).
Notice that the green ellipse intersects the blue circle at (r,r).

So the green ellipse is x^2/a^2 + y^2/L^2 = 1, and all we want to do is find "a" in terms of "r" and "L".

r^2/a^2 + r^2/L^2 = 1 => a^2 = (r^2L^2)/(r^2-L^2)

The elliptical curve traced by the red dot in the previous image is a segment of the green ellipse given by

x^2/((r^2L^2)/(r^2-L^2)) + y^2/L^2 = 1,

where "r" and "L" are the previously described constants.

and from there I could easily find the y(x) describing the segment of the ellipse that is the curve that I was looking for originally.

10. Oops!

I made a mistake. Luckily I caught it before anyone else did, that would have been a bit embarassing.

In my previous post, wherever I had written "r^2 - L^2", it should be reversed and written as "L^2 - r^2",
since if it is computed using the former then the resulting curve will be concave up curve, and definitely not what we are looking for.

11. Originally Posted by Luc
OK so I threw calculus out the window and just solved it with geometry after I realized that the curve I was trying to find was elliptical (thanks to the article that mr.fantastic posted).
It was alot easier.

I'll just walk you through what I did, using the attached picture.

I called the constant distance along the curve from the previous image "L", and made (0,L) the vertice of the ellipse who's identity we are finding.
I kept the circle that I started with in the other picture and made it of radius "r", centred at (r,0).
Notice that the green ellipse intersects the blue circle at (r,r).

So the green ellipse is x^2/a^2 + y^2/L^2 = 1, and all we want to do is find "a" in terms of "r" and "L".

r^2/a^2 + r^2/L^2 = 1 => a^2 = (r^2L^2)/(r^2-L^2)

The elliptical curve traced by the red dot in the previous image is a segment of the green ellipse given by

x^2/((r^2L^2)/(r^2-L^2)) + y^2/L^2 = 1,

where "r" and "L" are the previously described constants.

and from there I could easily find the y(x) describing the segment of the ellipse that is the curve that I was looking for originally.
Hmmmm ..... as the height of the ellipse is allowed to go to zero shouldn't the locus smoothly continue from its point on the circle ......? In which case, the x-intercept of the locus should be (L, 0) ......?

This would mean that if the locus is an ellipse, this ellipse will have an axis that lies on the line y = x (its point on the circle will therefore be a vertex) .... So I'm not sure that x^2/a^2 + y^2/L^2 = 1 is the correct model for the locus .....

Perhaps a dynamic geometry software could be used to shed more light on this ....?

12. Another idea, probably better suited to the original problem:

Find it parametrically.
The family of ellipses looks like this:

x^2 + (y^2/t) = 1

Centre these at the origin, and t ranges from 1 to infinity. 1 is the unit circle of course. The red dot is (-1,0), and the first blue dot is (0,1). The distance between these two points is the square root of 2. So then any point on our little curve must satisfy the equations:

x^2 + (y^2/t) = 1

(x-(-1))^2 + y^2 = 2

You can multiply the first equation by t, and then subtract the two equations in order to solve for x in terms of t. You get x=-1+t, after using the quadratic formula. Then you get y in terms of t.