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From: gudi on 10 Apr 2010 02:19
On Apr 9, 3:12 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
> Thomas Nordhaus schrieb:
> > (Result): Provided that
> > (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a

Trying to follow this. Letting c^2 = a^2 - b^2 and C^2 = A^2 - B^2...

For the special case of smaller circle c = 0 , B/a > Infinity
and the other special case of bigger circle C = 0 , A/a < 0
OK? !

> > the minimal distance D between e1 and e2 is given by:
> > (2) D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
> Correction:
> D^2 = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
> --
> Thomas Nordhaus

If either c or C is zero, then the circle should touch the ellipse?
Please check.

Narasimham
From: Thomas Nordhaus on 10 Apr 2010 04:22
gudi schrieb:
> On Apr 8, 4:47 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
>>
>> One can get this result by computing the normal vectors n1, n2 to e1 and
>> e2 at points P, Q and use the fact that n1,n2 and P-Q must be collinear.
>
> For this situation with common normal PQ, the condition of parallelism
> of n1 and n2 or tangents t1 and t2 alone is sufficient.

But we have to find those points with common normal first. And that's
where I use *two* conditions: (1) n1 collinear n2, and (2) n1 collinear
(Q-P). Collinearity of the normal vectors alone does not determine P and
Q uniquely. In fact - for every P in e1 there is a Q in e2 such that
n(P) is collinear to n(Q).

--
Thomas Nordhaus
From: Thomas Nordhaus on 10 Apr 2010 04:46
gudi schrieb:
> On Apr 9, 3:12 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
>
>> Thomas Nordhaus schrieb:
>
>>> (Result): Provided that
>>> (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a
>
> Trying to follow this. Letting c^2 = a^2 - b^2 and C^2 = A^2 - B^2.

OK.

>
> For the special case of smaller circle c = 0 , B/a > Infinity
> and the other special case of bigger circle C = 0 , A/a < 0
>
> OK? !

The assumption was that a>b, A>b. But if you fix A,B and b and let a->b
then the middle-term of the inequality becomes bigger then B/a and the
inequalty becomes void which means the formula for D cannot be applied.
In that case the minimum for D will either be A-a or B-b.


>
>>> the minimal distance D between e1 and e2 is given by:
>>> (2) D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>> Correction:
>>
>> D^2 = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>
> Needs one more correction to D^6, to tally equation dimension?

I dont understand - the dimensions are correct: D^2 is of order two. The
r.h.s is of order 2*2/2 = 2. That checks.

>
> If either c or C is zero, then the circle should touch the ellipse?
> Please check.

Mhhm - I think you're overcomplicating matters. The case you stated
isn't covered by the assumptions. The formula for D doesn't work in
this case(isn't even defined). What is true is the following: If the
assumptions on a,b,A,B are'nt met, then

--
Thomas Nordhaus
From: Thomas Nordhaus on 10 Apr 2010 05:04
Thomas Nordhaus schrieb:
> Mhhm - I think you're overcomplicating matters. The case you stated
> isn't covered by the assumptions. The formula for D doesn't work in
> this case(isn't even defined). What is true is the following: If the
> assumptions on a,b,A,B are'nt met, then

Oops! ... then D = min{A-a,B-b}.


--
Thomas Nordhaus
From: Robert H. Lewis on 12 Apr 2010 18:30
>
> (Problem): Given ellipses e1: x^2/a^2 + y^2/b2 = 1
> and e2: x^2/A^2 + y^2/B^2 =1. Compute the minimal distance D between
> points P, Q lying in e1, e2 resp.
>...
> Here is what I found out:
>
> (Result): Provided that
> (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a
>
> the minimal distance D between e1 and e2 is given by:
>
> (2) D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>
> For example let a=2, b=1, A=2.5, B=1.5 then D ~=
> 0.47871. Compare that to (A-a)=(B-b)=0.5. ...

I think (1) was supposed to be (A/a) < (A^2-B^2)/(a^2-b^2) < B/b.

This is really cool, thanks for sharing.

I thought of generalizing it. Let f1 be an ellipse centered at the origin and f2 any conic. We can write their equations as
f1 := a1^2*x1^2 + b1^2*y1^2 - 1;
f2 := a2*x2^2 + b2*y2^2 + c2*x2 + d2*y2 + e2*x2*y2 + ff;

Then there are four equations, based on the equations and partial derivatives:

f1 = 0
f2 = 0
Deriv(f1,x1,1)*Deriv(f2,y2,1) - Deriv(f1,y1,1)*Deriv(f2,x2,1) = 0
(y2 - y1)*Deriv(f1,x1,1) - (x2 - x1)*Deriv(f1,y1,1) = 0

A resultant can be computed for each of x1, x2, y1, y2 in turn. Numerical values could then be plugged in for the parameters, and the various possibilities checked.

Using the Dixon resultant, I got the resultant for x1 to have 1529 terms and the one for x2 to have 10400 terms. Each takes a few seconds to compute.

Alternatively, one can work with ds =distance between the points (x1,y1) and (x2,y2). There are then five equations, and one can eliminate x1, y1, x2, y2 and have a resultant for ds in terms of the parameters. That is a lot harder, and blows out my RAM.

Robert H. Lewis
Fordham University
http://fordham.academia.edu/RobertLewis
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