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Question: Prove that a similarity transformation (replacing $x$ by $tx$ and $y$ by $ty$) carries an ellipse with center at the origin into another ellipse with the same eccentricity.

(The next questions go on to prove the same result (and converses) for hyperbolas.)

Please don't feel the need to read through all my attempts, I just wanted to show what I had tried and give some background on what has been covered in the book.

I think my question boils down to this: am I allowed to say "given some conic, align the coordinate axes so that the origin is at the center of the conic and the x-axis is aligned with the main axes of the conic" without any further justification?


My Attempt

This is easy to prove for an ellipse (or hyperbola) in standard Cartesian form:

$$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$$

Replacing $x$ and $y$ by $tx$ and $ty$ respectively, we have

$$\frac{(tx)^2}{a^2}+\frac{(ty)^2}{a^2(1-e^2)}=\frac{x^2}{\left(\frac a t\right)^2}+\frac{(y)^2}{\left(\frac a t\right)^2(1-e^2)}=1$$

Which which is another ellipse with a center at the origin with the same eccentricity. The problem is that the standard form only describes ellipses and hyperbolas with vertical directrices. Intuitively I understand that you could choose the coordinate axes so that any ellipse is described by this equation, but this has not been made explicit in the book.

I suppose one question worth asking, at this point, is if this is all Apostol wanted me to prove? Taking the original question literally, it seems my argument for choosing the coordinate axes is a bit of hand-waving to get at the other cases of ellipses.

EDIT: After discussing this with a professor at the university I attended, his interpretation was that Apostol's (high-level) intention was just to show that scaling the shape of an ellipse does not change the eccentricity, because eccentricity is a description of the shape and not the size. He thought, after reading the chapter, that I should consider only those curves in the ideal standard form for all questions such as this, and if necessary remark (although it is yet unproven in the book) that any conic is congruent to one in standard form. In this case, my original solution is adequate.

EDIT #2: It now seems clear that Apostol's intention was to consider only the standard cases, as the next question is Use the Cartesian equation which represents all conics of eccentricity $e$ and center at the origin to prove that these conics are integral curves of the differential equation $y'=(e^2-1)x/y$. Indeed, it seems that the integral curves of that differential equation are given by $\frac{x^2}{C}+\frac{y^2}{C(1-e^2)}=1$, which are really only the central conics at the origin with horizontal directrices, not all conics at the origin. Since, in that question, he does not make a distinction, I think he is just being atypically loose with his terminology.

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Note that the next question is "If two concentric ellipses have the same eccentricity and major axes on the same line, then they are related by a similarity transformation.". Perhaps Apostol does expect me to just accept a bit of hand waving and say that I place the origin of my coordinate axes on the center of the ellipses, and the x axis in the direction of the major axes. If this is acceptable, it certainly speeds up the proof. –  Michael Boratko Oct 26 '11 at 21:53
    
This doesn't have to do with your question, but one of the earliest and cruelest jokes of coordinate geometry is how replacing $x$ and $y$ with $tx$ and $ty$ in an equation for a curve does not give the equation for the image of that curve under the transformation of $\mathbb{R}^2$ given by $(u,v) \mapsto (tu, tv)$. (Try the unit circle and $t=2$. One must replace $x$ and $y$ with formulas for the inverse transformation.) For your problem, the result follows if you can prove that any ellipse can be rotated to one that your argument applies to, and that rotation does not change eccentricity. –  leslie townes Oct 26 '11 at 22:02
    
That's interesting. The idea of rotation hasn't really been broached in the book yet, although I see how that proves my argument. One of the next questions after this is Use the Cartesian equation which represents all conics of eccentricity $e$ and center at the origin to prove that these conics are integral curves of the differential equation $y'=(e^2-1)x/y$. This is easy if we're allowed to start by saying "Take the coordinate axes to have origin at the center of the conic, and the x axis parallel to the main axes of the conic." Otherwise I'm not sure what the Cartesian equation would be. –  Michael Boratko Oct 26 '11 at 22:10
    
Here are two approaches. Since I suspect neither suits your needs, it will have to be done in comments. (a) The eccentricity is defined as a ratio. The major, minor axes are stretched by same factor. So eccentricity is unchanged. (b) Transform, Solve, Transform Back. Specifically, rotate about the center until major, minor axes are in the standard place. Eccentricity does not change. Dilate by factor $t$. You showed eccentricity does not change. Rotate back. Eccentricity does not change. –  André Nicolas Oct 26 '11 at 22:21
    
@AndréNicolas While I understand both approaches, I agree that they probably won't work for me in this problem if I want to stay on course with the book (which I do). I'm starting to grow more comfortable with my "generalized attempt", can you comment on whether it is definitely wrong? –  Michael Boratko Oct 26 '11 at 22:27
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3 Answers

up vote 2 down vote accepted

A central conic with center at the origin takes the general form

$$ax^2+bxy+cy^2=1$$

(missing linear terms; why?). We are being asked to prove that

$$t^2(ax^2+bxy+cy^2)=1$$

also has the same eccentricity.

The rotation that will zero out the cross-term will be the same for both, a fair bit of algebra shows that for the rotation

$$\begin{align*}x&=p\cos\,\varphi-q\sin\,\varphi\\y&=p\sin\,\varphi+q\cos\,\varphi\end{align*}$$

taking the $\varphi$ given by

$$\tan\;\varphi=\frac{b}{a-c+\sqrt{b^2+(a-c)^2}}$$

does the trick.

Applying that transformation yields the conic

$$\frac{a+c+\sqrt{(a-c)^2+b^2}}{2}p^2+\frac{a+c-\sqrt{(a-c)^2+b^2}}{2}q^2=1$$

in the "unscaled" case, and with the pendant factor $t^2$ in the scaled case, which now looks a lot like the standard form you're accustomed to...

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From the linear algebra viewpoint, what we have actually done here is a special case of Jacobi's method for diagonalizing a symmetric matrix... –  J. M. Oct 26 '11 at 22:57
    
Thanks, this is obviously correct. I'm still pretty sure that this isn't the method that the book was looking for, normally he leads the proofs a little more if they require this many jumps from currently developed material, but I will go with it for now. –  Michael Boratko Oct 26 '11 at 23:10
    
At the very least, there is an easy way to derive the $\tan\,\varphi$ expression I gave: substitute the rotation transformations into the original conic, and equate the coefficient of the cross-term $pq$ to $0$. This yields an expression for $\tan\,2\varphi$; the double-angle formula + judicious use of the quadratic equation yields the $\tan\,\varphi$ expression. –  J. M. Oct 26 '11 at 23:21
    
That's true, but the two big steps are - he has not introduced rotations yet (at all), and he has not had us show that a central conic with center at the origin takes that form (missing the linear terms), which I think he would have had as a separate question. I do really appreciate your answer though. –  Michael Boratko Oct 26 '11 at 23:49
    
I did take your points. (apologies if it did not seem that way) :) I'll see if I can offer up a neater way that what I know to show that a conic with linear terms missing has to be central; the proof I am accustomed to will double the length of this answer... –  J. M. Oct 26 '11 at 23:53
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The general Equation of a ellipse whose axes are parallel to $X-Y$ axes is

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ and the eccentricity is given by $e=\sqrt{1-\dfrac{b^2}{a^2}}$ (I am assuming $a>b$).

If a similarity transformation occurs i.e, $(x,y)\to(tx,ty)$, then new equation of ellipse is $$\dfrac{(tx)^2}{a^2}+\dfrac{(ty)^2}{b^2}=1\ or \dfrac{x^2}{(a/t)^2}+\dfrac{y^2}{(b/t)^2}=1$$ So the eccentrity of the new ellipse $e'=\sqrt{1-\dfrac{(b/t)^2}{(a/t)^2}}=e$

In case Hyperbola just replace $b^2$ with $-b^2$ and remove the condition $(a>b)$.

Hence proved.

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This is my first attempt above. However, this doesn't address ellipses whose directrices are not parallel to the coordinate axes. –  Michael Boratko Oct 27 '11 at 11:56
    
You can always rotate the axes and write the can write thet equation in standard form. –  Ramana Venkata Oct 27 '11 at 11:59
    
Yes, intuitively that makes sense to me. The question is related to rigorously proving that fact, or (at the very least) rigorously proving that the eccentricity doesn't change due to rotation. –  Michael Boratko Oct 27 '11 at 12:23
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my question boils down to this: am I allowed to say "given some conic, align the coordinate axes so that the origin is at the center of the conic and the x-axis is aligned with the main axes of the conic" without any further justification?

If eccentricity is invariant under the operation of "align the coordinate axes so that ..." then calculations involving the eccentricity will not be affected by the choice of a conveniently aligned coordinate system. The invariance can be seen by defining eccentricity in terms that are invariant and do not refer to position in a coordinate system. For that it is enough to define major and minor axis invariantly.

The major axis is the longest chord of the ellipse, and the minor axis is the longest chord perpendicular to the major axis. The lengths of these axes are invariant under rigid motion or reflection of the plane. The ratio of major to minor axes is also invariant under similarity transformations such as $(x,y) \to (tx,ty)$. The eccentricity can be calculated from this ratio, so that it is both scale-invariant (because the ratio is) and isometry-invariant (because it is a function of the axes).

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The statement - "The lengths of these axes are invariant under rigid motion or reflection of the plane" - seems to be difficult to prove using the tools so far introduced. In particular, concepts such as "rigid motion" and "reflection" have not yet been introduced. –  Michael Boratko Oct 26 '11 at 22:44
    
For this question it is only necessary to check that all distances change by a factor of $t$ under the similarity transformation, which can be done using the formula for distance. –  zyx Oct 27 '11 at 0:26
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