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Does any ellipse $E$ have a circle $C$ such that you can obtain $E$ by transforming $C$ by a simple formula $F$? In details , both $E$ and $C$ have the same center and the axes of $E$ are the XY axes. And F moves $(x,y)$ to $( m*x , y)$ . Where m is a real number.

Thank you in advance.

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Yes, but the formula is of the specific kind you mention only after rotating the ellipse and centering it at the origin. – ShreevatsaR Nov 7 '11 at 12:58
Sure, if you scale a circle differently in the horizontal and vertical directions, you'll certainly get an ellipse. – J. M. Nov 7 '11 at 12:59
@J.M.: Of course, it's enough to scale (a different circle) only horizontally. :-) – ShreevatsaR Nov 9 '11 at 3:08
up vote 4 down vote accepted

I think you'll see it from $$ {\rm ellipse:\ \ }{x^2\over a^2}+{y^2\over b^2}=1. $$

$$ {\rm circle:\ \ } {x^2\over b^2}+{y^2\over b^2}=1. $$

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That formula for an ellipse is only valid for axis-parallel ellipses centered at the origin. Anyway, to complete it into the form the OP wants: $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ means $\displaystyle \frac{b^2}{a^2}x^2 + y^2 = b^2$, which is the circle $x'^2 + y^2 = b^2$ with $(x',y)$ sent to $(\frac{a}{b}x', y)$. – ShreevatsaR Nov 7 '11 at 13:22

Errata: For some reason, the answer I just submitted did not print the end of one sentence and the last paragraph. The entire sentence read: "In addition, if the semi-major axis is allowed to remain length 1, a=1 and 0

If you Google "circle transformation ellipse" you will see that Archimedes explored the transformation. Yet, even our new Commoon Core does not mention the close connections that conic sections have to other simple mathematical concepts.

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errata explained: I guess the website does not allow the use of the inequality symbols. The omitted end of the sentences read 0 is less than b is less than 1 which suggests that we let $b=sin(\theta)$. The foci are at $\pm cos(\theta)$, the directrices at $\pm sec(\theta)$ and the semi-focal width is $sin^{2}(\theta)$.

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I've gone ahead and edited your posts to add MathJax formatting. This site supports a subset of LaTeX. You can read more on it here:… – ml0105 Apr 6 '14 at 17:04

Since $\displaystyle \frac{x}{a}$ accomplishes a horizontal shear of magnitude $a$ and similarly for $\displaystyle \frac{y}{b}$ in the vertical direction, the unit circle $x^2 + y^2 =1$ after the shear will be the ellipse:

$$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} = 1$$

In addition, if the semi-major axis is allowed to remain length $1$, $a=1$ and $0<b<1$. The latter suggests $b$ can be represented by $\sin(\theta)$. Explore and you will find that this unit ellipse has foci at $\pm \cos(\theta)$, directrices at $\pm \sec(\theta)$ and focal width $2\sin^{2}(\theta)$.

The connections between the unit circle, transformations and trigonometry are important for secondary students to understand. It is somewhat typical of textbooks to introduce a topic like conic sections with formulas of their own as if they were a different species of mathematical functions.

Even our new Common Core does not use the approach. If you Google "circle transformation ellipse" you will see that Archimedes noted this connection.

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The last two paragraphs belong on Math Educators (stack exchange site), not here. Please edit your post accordingly. – amWhy Apr 6 '14 at 16:48

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