From: Thomas Nordhaus on
Hi,
I'm looking for an "elegant" geometric proof of of a result I and others
worked out completely algebraically and by using CAS. The equations
turned real messy quick but at the end everything turned out nice. So I
suspect some deeper lying and geometric arguments exist.

(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.

W.l.o.g assume that a>b, A>a, B>b. So e2 lies on the outside of e1.
Candidates for points P and Q where this minimum is attained are
adjacent vertices of the ellipses. That would mean D = A-a or D=B-b
whatever is smallest. However this is not always true. In some cases the
minimal distance is attained "in between" vertices. 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. One can easily see from (1) that for example similar
and confocal ellipses do not meet condition 1. Given a and b (1) yields
a tongueshaped region in (A,B) space.

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.

Question: Is there a geometric proof to this without all these messy
computations?

--
Thomas Nordhaus
From: gudi on
On Apr 8, 4:47 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
> Hi,
> I'm looking for an "elegant" geometric proof of of a result I and others
> worked out completely algebraically and by using CAS. The equations
> turned real messy quick but at the end everything turned out nice. So I
> suspect some deeper lying and geometric arguments exist.
>
> (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.
>
> W.l.o.g assume that a>b, A>a, B>b. So e2 lies on the outside of e1.
> Candidates for points P and Q where this minimum is attained are
> adjacent vertices of the ellipses. That would mean D = A-a or D=B-b
> whatever is smallest. However this is not always true. In some cases the
> minimal distance is attained "in between" vertices. 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. One can easily see from (1) that for example similar
> and confocal ellipses do not meet condition 1. Given a and b (1) yields
> a tongueshaped region in (A,B) space.
>
> 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.

Narasimham

> Question: Is there a geometric proof to this without all these messy
> computations?
>
> --
> Thomas Nordhaus

From: gudi on
On Apr 8, 4:47 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
> Hi,
> I'm looking for an "elegant" geometric proof of of a result I and others
> worked out completely algebraically and by using CAS. The equations
> turned real messy quick but at the end everything turned out nice. So I
> suspect some deeper lying and geometric arguments exist.
>
> (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.
>
> W.l.o.g assume that a>b, A>a, B>b. So e2 lies on the outside of e1.
> Candidates for points P and Q where this minimum is attained are
> adjacent vertices of the ellipses. That would mean D = A-a or D=B-b
> whatever is smallest. However this is not always true. In some cases the
> minimal distance is attained "in between" vertices. 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. One can easily see from (1) that for example similar
> and confocal ellipses do not meet condition 1. Given a and b (1) yields
> a tongueshaped region in (A,B) space.
>
> 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.
>
> Question: Is there a geometric proof to this without all these messy
> computations?
>
> --
> Thomas Nordhaus

Perhaps for this situation with common normal PQ, the condition of
parallelism of n1 and n2 or tangents t1 and t2 alone is sufficient,
including all the possibilities comprehensively.

Narasimham
From: Thomas Nordhaus on
Thomas Nordhaus schrieb:
....
> (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))

Correction:

D^2 = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))

--
Thomas Nordhaus
From: gudi on
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

Needs one more correction to D^6, to tally equation dimension?

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

Narasimham