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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 ?

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  • $\begingroup$ The orthogonal projection is onto the line carrying the axis of length $2b$. The geometric issue at stake here is absolutely obvious, and is completely covered in Andrew D. Hwang's answer. $\endgroup$ – Christian Blatter Dec 31 '15 at 19:21
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$\newcommand{\Reals}{\mathbf{R}}$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 $$ \frac{u^{2}}{b^{2}} + \frac{v^{2}}{b^{2}} \leq 1,\quad\text{or}\quad u^{2} + v^{2} \leq b^{2}. \tag{2} $$

Stretching a disk to an ellipse

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".)

Orthogonal projection of an ellipse to a disk

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  • $\begingroup$ This shows that the area of an ellipse is $\pi ab$ but how can I know that there exists an orthogonal projection such that $(u, v) \mapsto (x, y) = (\frac{a}{b}u, v) $in the first place ?My main concern here is about the existence of such ort. project . $\endgroup$ – Nameless Jan 1 '16 at 10:48
  • $\begingroup$ The term "orthogonal projection" usually entails projecting to a space of lower dimension, e.g., representing the ellipse as a slanted section of a cylinder in space and projecting along the axis of the cylinder, as in coffeemath's answer. (You can accomplish such a mapping by closing one eye and placing your head to the left or right of your screen so that the right-hand ellipse looks circular.) Conceivably, however, your author meant "orthogonal stretching", perhaps in the sense of looking at an ellipse at an angle to see a circle. Does your book say what "orthogonal projection" means? $\endgroup$ – Andrew D. Hwang Jan 1 '16 at 13:14
  • $\begingroup$ My main difficulty here is about accepting the fact that there exists such a way to cut off a section of a cylinder in space such that its orthogonal projection along its axis gives me a transformation map as the one you've defined. That's the crucial point I am really trying to prove to my self.About your last question:my book defines an ortogonal projection as the shadow of an object formed if the sun would be directly upon it (which I can understand is the same as a projection of a section of a cylinder along its axis) (contin...) $\endgroup$ – Nameless Jan 1 '16 at 13:37
  • $\begingroup$ (ued...) I would like to see the proof of the existence of such projection mathematically not intuitively,i.e roughly speaking if I would play with an ellipse and its shadow ,when the sun is over it ,is there a projection as the one you've defined ? $\endgroup$ – Nameless Jan 1 '16 at 13:38
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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.

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  • $\begingroup$ I've got a doubt ,if the distance from the origin to where the plane cuts the the vertical plane $y=a$ is $b$,then don't we have that the radius of our circle is $\sqrt{b^2-a^2} $ ?(The cylinder I am visualizing is centered at the origin ) $\endgroup$ – Nameless Nov 13 '15 at 9:05
  • $\begingroup$ @Nameless Actually that radical is the vertical distance from the point $(0,b,0)$ to the intersection of the tilted plane with the ellipse. The circle is still of radius $a$ and lies in the plane $z=0.$ $\endgroup$ – coffeemath Nov 13 '15 at 13:00
  • $\begingroup$ @Nameless In last comment, vertical distance from $(0,b,0)$ to where the tilted plane intersected with the vertical plane $x=0$ meets the ellipse, I should have said. $\endgroup$ – coffeemath Nov 13 '15 at 13:23
  • $\begingroup$ You say that the circle is still of radius $a$,shouldn't it be of radius $b$ ?I am confused. $\endgroup$ – Nameless Nov 13 '15 at 13:25
  • $\begingroup$ Can one prove the above claim by some geometric continuity argument ? $\endgroup$ – Nameless Nov 13 '15 at 13:32
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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

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