# FOM: conics

Neil Tennant neilt at mercutio.cohums.ohio-state.edu
Thu Mar 1 12:29:28 EST 2001

```On Wed, 28 Feb 2001, Matthew Frank wrote:

> I can't help with references off-hand, but here's an elementary proof:
>
> Let F and G be the foci of the ellipse.
> Let P be a point on the ellipse.
> Let APB be the tangent to the ellipse at P.
>
> Then of all points P' on AB, P is the one which minimizes the distance FP'
> + P'G (because all the other points on the line are outside the ellipse).
>
> Let H be the reflection of G about the line AB.
> Then for any point P' on AB, P'G = P'H.
******
> So P is also the point on AB which minimizes the distance FP' + P'H.
> Hence P is at the intersection of FH with AB.
######
> So angle APF = angle HPB = angle BPG.
> QED.

Thanks. I had found this proof earlier (on the web!) and showed it to my
colleague Lisa Shabel who specializes in Kant's philosophy of geometry,
and who knows her Euclid. Her immediate reaction was that from the point
marked ****** on, the proof appeared to be somewhat "un-elementary". I
can't tell what it may be in the logical structure of the reasoning *as it
stands* that would prompt such a reaction, but I think I can make the
reasoning appear a little more elementary as follows. (My method appears
to avoid the notion of actually attaining the minimization in question.)

For the section ****** to ###### of the proof above, reason instead as
follows:

******
By definition of an ellipse:

for all I on the ellipse FI+IG is some constant, say k.

Let FH intersect the tangent AB in E.

Assume for reductio ad absurdum that E is distinct from the point of
tangency P.

Since E is on the tangent, but is not the point of tangency with the
ellipse, E lies in the exterior to the ellipse.
Hence FE+EG > k
Now EH=EG (by construction of H).
So FE+EH > k. .......(1)

By choice of E, FE+EH = FH.
By the triangle inequality, FH < FP+PH.
So FE+EH < FP+PH.
But PH = PG, by construction of H.
So FE+EH < FP+PG.
But FP+PG = k, since P is on the ellipse.
So FE+EH < k, contrary to (1).

Therefore E is identical to the point of tangency P; i.e. P is at the
intersection of FH with AB.
######

We assumed not(E=P), derived a contradiction, and concluded E=P.
Do intuitionist geometers allow this?

Neil Tennant

```