Never heard of this 'OBLONG METHOD' before, but i am assuming that it means a method by which you can construct an ellipse inside a given rectangle such that the sides of the rectangle are tangents to the ellipse and parallel to the major and minor axes, which is just a very long round-about way of saying that we want that ellipse which is bounded by the given rectangle.
Seemed like an interesting problem, so i gave it a shot and came up with this answer, which is pretty cool, so i will state it here :

Let the rectangle be ABCD, with ab || cd
m,n are mid-points of AD and BC resp.

A--------------------- B
| |
| |
m n
| |
| |
D--------------------- C-----------------E

Extend DC to the right to a point E such that BE = AB.

Then CE is the distance between the foci of the required ellipse.
Get the centre of rect ABCD as O, and place CE parallel to CD (along MN) such that the mid-point of CE coincides with O. The point on the left is one focus F1, and the one on the right is the other focus F2.

So, now we know where the two foci are, measure out 2F1M + F1F2.
Call this distance as d.

Draw with F as centre a circle with radius d.
( either f1 or f2,anyone, in the rest that follows,make f(1,2) = f'(2,1) )

Take any point P on the circle. Join PF and PF'. Draw the perpendicular bisector of PF'. The Point E where it cuts the seg. PF is a point on the required ellipse.
Take different P's to get as many E's as you want.

I am sure after doing this on a piece of paper, the method behind this madness will be readily comprehendible, with almost none or very little thought, but that " Damn! That was so simple" reaction is like applause to a performer.

I hope Dr. Alex B. will not be cut off this post for the poor drawing attempt. Especially since this site has got so many cool funky moving JAVA applets...

Q.E.D.
:wq