Problem: Prove that the area of an ellipse with major axis and minor axis of lengths $2a$ and $2b$,respectively, is $ab \pi$ .
Proof: We do this by projecting the ellipse into a figure whose area we can find,namely a circle with diamenter $2b$. To take advantage of
the costant ratio of area,we must have some other relevant figure
projected along with the ellipse. For this example,we consider the
triangle formed by the endpoints of the major axis and one endpoint of
the minor axis. Let this triangle be $ABC$ and its projection be
$A'B'C'$. Hence we have $[ABC]=ab$ and $[A'B'C']= b^2$ .Since
orthogonal projections preserve ratio of area,we have $$ \cfrac {[ Ellipse]} { [ABC]}= \cfrac { [ circle ]}{[A'B'C']} $$ where [Ellipse]
and [Circle] are the areas of the ellipse and the circle .
(…)
hence area of ellipse $= ab \pi$.
Question: I've tried hard to visualize what the author is telling,but I can't really see how we can actually project orthogonally an ellipse such that we have a circle with diameter $2b$.
I've tried to make such projection with Geogebra,but,while I can make a circle of radius $b$ by projecting the endpoints of the minor axis of the ellipse,I still fail to have any other point of the ellipse on this circle…
So my question is :what is a valid argument to prove a priori the existence of such circle ?
Best Answer
$\newcommand{\Reals}{\mathbf{R}}$If you're willing to admit coordinate descriptions, an axis-aligned ellipse centered at the origin can be expressed as the solution set of the inequality $$ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1. \tag{1} $$ Under the "horizontal stretching" $(\frac{b}{a}x, y) = (u, v)$, the ellipse corresponds to the disk of radius $b$ defined by $$ \frac{u^{2}}{b^{2}} + \frac{v^{2}}{b^{2}} \leq 1,\quad\text{or}\quad u^{2} + v^{2} \leq b^{2}. \tag{2} $$
Since the transformation $(u, v) \mapsto (x, y) = (\frac{a}{b}u, v)$ scales areas (of axis-oriented rectangles, and hence of all measurable regions) by a factor of $\frac{a}{b}$, the disk of area $\pi b^{2}$ maps to the ellipse of area $\pi b^{2}(\frac{a}{b}) = \pi ab$.
Added in edit: Let $(u, v, z)$ denote Cartesian coordinates in $\Reals^{3}$ (preserving the notation of the original answer), and define the orthogonal projection $\Pi:\Reals^{3} \to \Reals^{2}$ by $\Pi(u, v, z) = (u, v)$.
Fix real numbers $0 < b < a$, and let $\theta$ be the unique real number with $0 < \theta < \frac{\pi}{2}$ satisfying $\frac{b}{a} = \cos\theta$.
Define the mapping $i:\Reals^{2} \to \Reals^{3}$ by \begin{align*} i(x, y) &= (x\cos\theta, y, x\sin\theta) \tag{3} \\ &= (\tfrac{b}{a}x, y, \tfrac{1}{a}\sqrt{a^{2} - b^{2}} x). \tag{4} \end{align*} Viewing $(x, y)$ as Cartesian coordinates in the plane $i(\Reals^{2}) \subset \Reals^{3}$, the ellipse (1) is precisely the intersection of $i(\Reals^{2})$ and the (solid) cylinder $\{(u, v, z) : u^{2} + v^{2} \leq b^{2}\}$. (Equation (3) shows the mapping $i$ is a rigid motion, while (4) shows that the image of the ellipse satisfies $u^{2} + v^{2} \leq b^{2}$.) The orthogonal projection $\Pi$ maps this slanted ellipse to the disk (2). (In the diagram, the ellipse has been translated "upward" along the axis of the cylinder, for visual clarity, so the ellipse is disjoint from its "shadow".)