Is it possible to project orthogonally an ellipse with major and minor axes $2a$,$2b$ so that its image is a circle with diameter $2b$?

Question

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 ?

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$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 $\pi b^{2}(\frac{a}{b}) = \pi ab$.

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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".)

Answer 2

Take the cylinder $x^2+y^2=a^2$ (with $z$ arbitrary), and make a plane passing through the $x$ axis which tilts upward in such a way that the distance from the origin to where the plane cuts the vertical planes $y=\pm a$ is the $b$ of your major axis $2b.$

Now take two spheres of radius $a$ which are to be bounded by the cylinder, one below and one above your plane. Imagine moving these until they just touch the plane, and say the upper sphere touches at $F_1$ and the lower sphere touches at $F_2.$ Then $F_1,F_2$ will serve as the foci of the apparent ellipse which is the intersection of the tilted plane with the cylinder, and once using geometry it is shown that the length sums are constant it is shown the apparent ellipse is an ellipse in fact.

I saw this idea somewhere but applied to the intersection of a tilted plane with a cone, to show that intersection was an ellipse, and that proof was called something like the "ice cream cone" proof. I think it also works here. There are some pictures going with the ice cream proof which might show in this case more vividly how it works.

here is a page linking to the icecream proof.

Answer 3

Here's how to visualize it in GeoGebra:

http://web.geogebra.org/?command=a=2;b=3;x^2/a^2=1-y^2/b^2;x^2=b^2-y^2

Drag slider 'a' to see the ellipse transforming into the circle