Still Harder Temple Geometry Problems
239
y,v Q(u,v) a
P(x,y) a
x,u
O
Figure 6.44. A point Q(u, v) on a circle can be transformed into a point P(x, y) on an ellipse by making the transformation x = u and y = (b/a)v.
If, on the other hand, we define two new variables u and υ such that u = x and υ = (a/b)y, then this equation becomes simply u 2 + υ2 = a 2 ,
which we recognize as the equation of a circle of radius a centered on the origin. In this case, we can regard the ellipse as a circle that has been shrunk by an amount b/a in the y direction. As mentioned in problem 6, we have just made what modern mathematicians call an affine transformation. The idea of shrinking a circle along one axis into an ellipse is illustrated in figure 6.44 and is the central idea employed in the traditional solution to this problem. We first apply the same method that was used to find the area element in problems 22 and 23 of chapter 5. Focusing attention on the small triangle at the top of figure 6.45, in which we consider the short y,v Δv
Δ l'
a y Δl Δy O
x Δx
x,u
Figure 6.45. For small enough triangles, the segments Δl′ and Δl can be considered straight lines.