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This is not meant to be a formal proof, but I just wanted to know if this is a valid way of thinking about the area of an ellipse. It does assume knowledge of the area of a circle, but this can be proven without knowledge of the area of an ellipse. I also don't know how to include pictures, so please excuse that.

Draw an ellipse with semi-major axis $a$ and semi-minor axis $b$. Because $a$ and $b$ are both linear quantities (i.e. they have units of distance), $k=\frac{b}{a}$ is dimensionless. Hence, we can draw a new ellipse with a semi-major axis of $ka$ and a semi-minor axis of $b$. Because $b=ka$, this new ellipse is a circle of radius $b$, so it has an area of $$A_{circ}=\pi b^2.$$ Because $a$ and $b$ are both linear, the area of the first ellipse, $A$, can be expressed as the product of these two quantities and some constant. Also, because $A$ was only scaled by a factor of $k$ in one dimension to get $A_{circ}$, $$kA=A_{circ}.$$ Substituting, $$\left(\frac{b}{a}\right)A=\pi b^2$$ $$\therefore \boxed{A=\pi ab}$$ as desired. $\blacksquare$

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    $\begingroup$ I don't know what you mean by "linear quantities". But the basic idea is sound: an ellipse with semi-major axis $a$ and semi-minor axis $b$ can be obtained from a circle of radius $b$ by scaling in one direction by a factor $a/b$. Since this transformation multiplies areas by $a/b$, the area of the ellipse is $a/b$ times the area of the circle. $\endgroup$ – Robert Israel Jun 11 '20 at 17:51
  • $\begingroup$ @RobertIsrael Thank you! By "linear quanities," I mean they have linear units, since they're in units of distance as opposed to square or cubic units. $\endgroup$ – Pendronator Jun 11 '20 at 18:00
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This is completely non rigorous, but $\pi ab$ is kind of the only thing it can be. If we find a formula for the area of an ellipse, it needs to meet a few conditions:

1: It only depends on $a$ and $b$

2: When either $a=0$, $b=0$, or both, $A=0$.

3: When $a=b$, the formula must reduce to $\pi a^2$

4: It must follow the typical properties of area: scaling one of the dimensions by a factor $k$ should produce an area scaled by a factor of $k$ and scaling both by a factor of $k$ should produce an area scaled by a factor of $k^2$.

5: It must be "symmetric" to $a$ and $b$, that is, interchanging them doesn't change the formula.

You can think long and hard about this, but $A=\pi ab$ is the only thing that works.

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    $\begingroup$ Thank you! It wasn't really meant to be rigorous. I was reading about conics in AoPS Volume 2, and this popped into my head. I just wanted to see if the foundations of the idea were sound. $\endgroup$ – Pendronator Jun 11 '20 at 18:02
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    $\begingroup$ Oh and another important criterion: the formula must be symmetric to $a$ and $b$, i.e, changing every $b$ to an $a$ and every $a$ to a $b$ shouldn't change the formula. $\endgroup$ – K.defaoite Jun 12 '20 at 15:14
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    $\begingroup$ nice answer....................... +1 $\endgroup$ – user766881 Jun 14 '20 at 1:13
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This is valid because an ellipse is indeed an affine transform of a circle. The hard part, of course, is proving that an ellipse is just a stretched circle, but that depends on how you define an ellipse in the first place.

A simpler, but equivalent, way to think about it is to just take a circle and stretch it until it becomes the ellipse. The ratio of the areas is just how much you stretched, which removes the need for proving $k$ is dimensionless and that type of stuff.

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The way you word your argument, it sounds like you're considering a family of ellipses, but you make inconsistent assumptions about which ellipses are in this family. At first you claim they all have the same $b/a$, but then you claim they include a circle, which for $a\ne b$ contradicts what you said originally.

However, that could be my misinterpreting what you meant when you said "$b/a$ is a constant". I've since concluded what you meant by "linear quantities" is quantities with the dimension of length, and by "constant" you meant "dimensionless". (Edit: your response to @RobertIsrael confirmed my interpretation of "linear", but I hadn't noticed it.)

Then your argument makes a lot more sense; you're seeing an ellipse as a circle scaled parallel to one axis, so its area varies in proportion to that axis. Then you're basically reasoning as @K.defaoite did.

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  • $\begingroup$ Thank you for your reply! You are correct; by "k=b/a is a constant," I meant that k was dimensionless. I'll edit my post to clear up the confusion. $\endgroup$ – Pendronator Jun 11 '20 at 18:29
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    $\begingroup$ @Pendronator It's an understandable mix-up: if $L$ is the length dimension a dimensionless quantity has dimension $L^0$, but if $L$ were a positive dimensionless number we'd have $L^0=1$. $\endgroup$ – J.G. Jun 11 '20 at 18:48

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