Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$

Question

Yes, I know this is the equation of an ellipse.

This is the starting point of the problem:

$$ r_1+r_2=2a, $$

where

$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$

I want to establish certain properties prior to deriving a parametric expression for the ellipse, or even the "standard form",

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. $$

I can do all of this with a couple tacks, a string and a pencil, but I want to show things algebraically. Preferably in a way that is directly applicable to all second-degree plane curves.

The first property I wish to establish is the range of $x$ values. I know from experience that the answer is going to be that the extreme values of $x$ will occur when $y=0$. The question that I am asking in this post is: what is the most direct (algebraic) way to show that $\left|x\right|$ is greatest when $y=0$?

I had hoped to establish that fact, and then use another (absolute value) method of determining the actual values of $x$ when $y=0$. But, the best I cam up with is this very long-winded demonstration.

First some definitions: Let $a,c,x,y\in\mathbb{R}$ where $a>c>0$ are constants, and $x$ and $y$ are variables such that, for $r_1\equiv\sqrt{(x+c)^2+y^2}$ and $r_2\equiv\sqrt{(x-c)^2+y^2}$ we have the equation

$$ r_1+r_2=2a. $$

The degenerate case has been omitted.

I will presume the greatest value to be $\left|x\right|=a$, and take the case of $x=a$. We square our equation

$$ (r_1+r_2)^2=4a^2 $$

$$ =r_1^2+2r_1 r_2+r_2^2 $$

$$ =(a+c)^2+y^2+2r_1 r_2+(a-c)^2+y^2 $$

$$ =2(a^2+c^2+y^2)+2r_1 r_2. $$

So

$$ r_1(2a-r_1)=a^2-c^2-y^2 $$

$$ =2a\sqrt{(a+c)^2+y^2}-((a+c)^2+y^2)=a^2-c^2-y^2. $$

Transpose, cancel and simplify

$$ 2a\sqrt{(a+c)^2+y^2}=2a^2+2ac, $$

$$ r_1=\sqrt{(a+c)^2+y^2}=a+c. $$

And, therefore

$$ r_2=2a-r_1=a-c. $$

Now, by definition we have

$$ \sqrt{\left(a+c\right)^2}\equiv\left|a+c\right|=r_1 $$

and

$$ \sqrt{(a-c)^2}\equiv\left|a-c\right|=r_2. $$

If $y\ne0$ while $x=a$, then

$$ r_1=\sqrt{(a+c)^2+y^2}=\left|a+c\right|+\delta_1, $$

and

$$ r_{2}=\sqrt{(a-c)^2+y^2}=\left|a-c\right|+\delta_2, $$

where both $\delta_1>0$ and $\delta_2>0.$

Since $a>c>0,$ we have $r_1+r_2=2a+\delta_1+\delta_2$, in contradiction to the equation we seek to satisfy. So $x=a\implies y=0.$

If $x>a$, let $\delta_3=x-a>0.$

So $$ r_1+r_2=a+c+\delta_3+a-c+\delta_3>2a; $$

another contradiction.

Thus far we have dertermined that $x=a\implies y=0,$ and for any $y$ that $x\le a$.

If we set $x=-a,$ then

$$ (r_1+r_2)^2=4a^2 $$

$$ =r_1^2+2r_1 r_2+r_2^2 $$

$$ =(-a+c)^2+y^2+2r_1 r_2+(-a-c)^2+y^2 $$

$$ =2(a^2+c^2+y^2)+2r_1 r_2 $$

$$ r_{1}\left(2a-r_{1}\right)=a^{2}-c^{2}-y^{2} $$

$$ 2a\sqrt{\left(-a+c\right)^{2}+y^{2}}-\left(\left(-a+c\right)^{2}+y^{2}\right)=a^{2}-c^{2}-y^{2} $$

$$ 2a\sqrt{\left(-a+c\right)^{2}+y^{2}}=2a^{2}-2ac $$

$$ r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c>0. $$

$$ r_{2}=2a-r_{1}=a+c. $$

Since we now have

$$ r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c, $$

and

$$ r_{2}=\sqrt{\left(a+c\right)^{2}+y^{2}}=a+c, $$

the argument showing that $x=-a\implies y=0,$ is essentially the same. Likewise for the argument that $x\ge-a.$

And that's just the $x$ range.

NB: I don't "need" any of this to fully characterize the ellipse. My goal is to find multiple ways of examining the same construct.

Answer 1

This is a constrained optimization problem that can be attacked directly with the method of Lagrange multipliers. Form the Lagrangian $x-\lambda(r_1+r_2-2a)$ and differentiate, producing the equations $$1-\lambda\left({x+c \over r_1}+{x-c \over r_2}\right) = 0 \\ -\lambda\left({y\over r_1}+{y\over r_2}\right) = 0.$$ The multiplier $\lambda$ cannot be zero and the two distances are always positive, so we must have $y=0$. Substituting this back into the equation $r_1+r_2=2a$ yields $x=\pm a$.

Strictly speaking, you should also show that $|x|$ can’t be greater than $a$, but that’s almost trivial.

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The above is one of the most straightforward solutions that I can think of, but if you insist on avoiding calculus, here’s another approach. The perimeter of the triangle formed by the foci and an arbitrary point on the ellipse is fixed at $r_1+r_2+2c=2(a+c)$. Using Heron's formula and the defining equation $r_1+r_2=2a$, the square of this triangle’s area can be expressed as a function of $r_1$: $$A^2 = (c-a+r_1)(c+a-r_1)(a+c)(a-c).\tag1$$ Since $a\gt c$, for this to be nonnegative we must have $(c-a+r_1)(c+a-r_1)\ge0$, or $$(r_1-a)^2\le c^2.\tag2$$ On the other hand, by construction every point on the ellipse is also the intersection of the circles $(x+c)^2+y^2=r_1^2$ and $(x-c)^2+y^2=r_2^2=(2a-r_1)^2$. From these equations we get $r_1=\frac ca x+a$. Substituting into (2) and simplifying results in $|x|\le a$. Plugging $\pm a$ into the defining equation of the ellipse and solving for $y$ yields $y=0$, as desired.

The bounds for $y$ can be found in a similar way. The $y$-coordinate of a point on the ellipse is the altitude of the triangle, so the maximum $y$ value will maximize the triangle’s area. Equation (1) is quadratic in $r_1$ and the $r_1^2$ term has a negative coefficient, so the value of $r_1$ that maximizes the area can be computed using standard algebraic techniques, and from that the corresponding values of $y$.

That wasn’t too bad, but I don’t really see much advantage to working directly with the radicals from the equation $r_1+r_2=2a$. It’s only a couple of more steps from $ar_1=cx+a^2$ to the standard equation of an ellipse. Once you have that equation, the ranges for $x$ and $y$ can be found by inspection, as Michael Hardy wrote in his now-deleted answer.

Answer 2

I am posting an answer. I am not claiming it is the best answer. It just shows an alternative approach to the problem. Part of why I found the problem difficult is that I find absolute-value arguments intimidating. Perhaps it's dyslexia, but these simple relations have always been difficult form me to process.

Once I actually considered the problem in these terms, I did see a fairly direct approach. What I am presenting here is a very laborious version, including my attempt at expressing things using quantifier logic. For the source of my unconventional quantifier notation, see https://mitpress.mit.edu/books/fundamentals-mathematics-volume-1

All of the following may be summarized by saying that once we realize $\left|x\right|\ge c\implies r_{1}+r_{2}=2\left|x\right|+\delta_{1}+\delta_{2}.$ the rest is easy.

First, a derivation using arbitrary truth-valued variables of a result used below.

$\left(p\wedge s\implies q\right)\wedge\left(p\wedge q\implies s\right)$

$\iff\left(\left(\neg\left(p\wedge s\right)\vee q\right)\wedge\left(\neg\left(p\wedge q\right)\vee s\right)\right)$

$\iff\left(\left(\neg p\vee\neg s\vee q\right)\wedge\left(\neg p\vee\neg q\vee s\right)\right)$

$\iff\left(\left(\neg p\vee\left(\neg s\vee q\right)\right)\wedge\left(\neg p\vee\left(\neg q\vee s\right)\right)\right)$

$\iff\left(\left(p\implies\left(s\implies q\right)\right)\wedge\left(p\implies\left(q\implies s\right)\right)\right)$

$\iff\left(p\implies\left(s\iff q\right)\right).$

Our topic will be the following: Given:$a,c,x,y\in\mathbb{R}$ where $a>c>0$ are constants, and $x$ and $y$ are variables such that, for

$$ r_{1}\left[x,y\right]\equiv\sqrt{\left(x+c\right)^{2}+y^{2}} $$ and

$$ r_{2}\left[x,y\right]\equiv\sqrt{\left(x-c\right)^{2}+y^{2}} $$ we shall investigate solutions to the definitive equation

$$ r_{1}+r_{2}=2a $$ of an ellipse.

Define the predicates:

$$ p\left[x,y\right]\equiv\left(r_{1}\left[x,y\right]+r_{2}\left[x,y\right]=2a\right), $$

$$ q\left[x,y\right]\equiv\left(\left|x\right|=a\right), $$

$$ r\left[x,y\right]\equiv\left(\left|x\right|<a\right), $$

$$ s\left[x,y\right]\equiv\left(y=0\right). $$

Theorem: For all $\mathfrak{p}=\left\{ x,y\right\} $ such that $r_{1}\left[\mathfrak{p}\right]+r_{2}\left[\mathfrak{p}\right]=2a$ the conditions $\left|x\right|\le a$ and $y=0\iff\left|x\right|=a$ hold.

If $\mathfrak{p_{\pm}}=\left\{ \pm a,0\right\} $ then $r_{1}\left[\mathfrak{p_{\pm}}\right]+r_{2}\left[\mathfrak{p_{\pm}}\right]=2a.$

Restated in symbolic form this becomes:

$$ \alpha)\underset{\mathfrak{p}}{\forall}\left(\left(p\wedge s\implies q\right)\wedge\left(p\wedge q\implies s\right)\right)\iff\left(p\implies\left(s\iff q\right)\right). $$

$$ \beta)\underset{\mathfrak{p}}{\forall}\left(q\wedge s\right)\implies p. $$

$$ \gamma)\underset{\mathfrak{p}}{\forall}p\implies\left(q\vee r\right). $$

Proof:

Define the utility functions

$$ \delta_{1}\left[x,y\right]\equiv r_{1}\left[x,y\right]-\left|x+c\right|, $$

$$ \delta_{2}\left[x,y\right]\equiv r_{2}\left[x,y\right]-\left|x-c\right|. $$

Without the argument lists and in expanded form, these are

$$ \delta_{1}\equiv\sqrt{\left(x+c\right)^{2}+y^{2}}-\sqrt{\left(x+c\right)^{2}}, $$

$$ \delta_{2}\equiv\sqrt{\left(x-c\right)^{2}+y^{2}}-\sqrt{\left(x-c\right)^{2}}, $$

Because $A>B>0\iff\sqrt{A}>\sqrt{B},$ it follows that $y=0\iff\delta_{1}=\delta_{2}=0,$ and for all $y$ that $\delta_{1}\ge0,$ $\delta_{2}\ge0.$

Let us examine the expressions $\left|x+c\right|$ and $\left|x-c\right|$ on the various intervals and boundaries relevant to the problem. These are,

$x<-a,$ $x=-a,$ $-a<x<-c,$ $x=-c,$ $-c<x<c,$ $x=c,$ $c<x<a,$ $x=a,$ $a<x.$

Why these intervals and values? We have proposed that $\left|x\right|>a$ holds no solutions for the definitive equation, and that $\left|x\right|=a$ are the only solutions when $y=0$, so $\left|x\right|<a$ holds no solutions when $y=0$. The graphs of the absolute value expressions will change slopes only at points where their arguments are zero.

All cases where $\left|x\right|\le c$ fall into two categories:

If $x\le-c$ then $x-c<x+c\le0,$ therefore $\left|x-c\right|=-\left(x-c\right)$ and $\left|x+c\right|=-\left(x+c\right).$

If $c\le x$ then $0\le x-c<x+c,$ therefore $\left|x-c\right|=x-c$ and $\left|x+c\right|=x+c.$

For $-x\le c$ we have $r_{1}+r_{2}=-\left(x+c\right)+\delta_{1}-\left(x-c\right)+\delta_{2}=-2x+\delta_{1}+\delta_{2}.$

For $c\le x$ we have $r_{1}+r_{2}=\left(x+c\right)+\delta_{1}+\left(x-c\right)+\delta_{2}=2x+\delta_{1}+\delta_{2}.$

Or, putting these results together,

$\left|x\right|\ge c\implies r_{1}+r_{2}=2\left|x\right|+\delta_{1}+\delta_{2}.$

Since $c<a,$ if $a\le\left|x\right|$then $r_{1}+r_{2}=2\left|x\right|+\delta_{1}+\delta_{2}\ge2a+\delta_{1}+\delta_{2}\ge2a$.

In this case, equality holds only when $\left|x\right|=a$ and $\delta_{1}=\delta_{2}=y=0.$

That is:

$r_{1}+r_{2}=2a\implies\left|x\right|\le a,$

$\left(r_{1}+r_{2}=2a\wedge\left|x\right|=a\right)\implies y=0,$

and,

$\left(\left|x\right|=a\wedge y=0\right)\implies r_{1}+r_{2}=2a.$

In terms of our predicates:

$\underset{\mathfrak{p}}{\forall}p\implies q\vee r,$ which is $\gamma$.

$\underset{\mathfrak{p}}{\forall}p\wedge q\implies s,$ which is half of $\alpha.$

$\underset{\mathfrak{p}}{\forall}q\wedge s\implies p,$ which is $\beta.$

It remains to be shown that $\underset{\mathfrak{p}}{\forall}p\wedge s\implies q.$

That is, to show that $\left(r_{1}+r_{2}=2a\wedge y=0\right)\implies\left|x\right|=a.$ For this, we assume $y=0$ and find all solutions under this condition. Since $\delta_{1}=\delta_{2}\iff y=0,$ our goal is to show there are no solutions for

$$ r_{1}+r_{2}=2a=\left|x+c\right|+\left|x-c\right|\wedge a>\left|x\right|. $$

The reason for the strict inequality is that we have already shown $r_{1}+r_{2}=2a\implies a\ge\left|x\right|,$ and $\left(r_{1}+r_{2}=2a\wedge a=\left|x\right|\right)\implies y=0.$

To this end it is instructive to examine plots of the absolute value expressions as functions of $x.$

The plots range from $x=-a$ to $x=a$. The solid red trace depicts values of $\left|x+c\right|$. It changes direction at $x=-c$. The dashed red trace depicts the continuuation of $x+c$. The solid blue trace depicts $\left|x-c\right|,$ which changes direction at $x=c$. The dashed black horizontal line represents the constant value $2a$. The magenta trace depicts the hypothetical $r_{1}+r_{2}=2a=\left|x+c\right|+\left|x-c\right|$. This reveals that the only places where $y=0$ satisfies the definitive equation are at $\left|x\right|=a$, which are not in the interval $a<\left|x\right|.$ It appears unnecessary to write out a formal demonstration of this finding for each part of $a<\left|x\right|.$

We have now demonstrated that $r_{1}+r_{2}=2a\wedge y=0\implies a=\left|x\right|,$

which is$\underset{\mathfrak{p}}{\forall}p\wedge s\implies q,$ the second half of

$$ \alpha)\underset{\mathfrak{p}}{\forall}\left(p\wedge s\implies q\right)\wedge\left(p\wedge q\implies s\right)=p\implies\left(s\iff q\right). $$

This completes the proof.

Answer 3

There is a very simple and short answer. Using the definition of monotonicity given in Fundamentals of Mathematics, Volume 1: Foundations of Mathematics: The Real Number System and Algebra Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle, which is a monotonic function is such that $$x<y\implies f\left(x\right)<f\left(y\right)$$ or such that $$x<y\implies f\left(y\right)<f\left(x\right),$$

and using the results of the elementary theory of numbers developed in Chapter 1B-1 for exponents, the proof is straight forward. That was the part I was missing. The result is "intuitively obvious", but I did not have a theorem to apply. In particular, I had no proof of the the blatantly obvious fact $$\sqrt{(x-c)^2+y^2}\ge\sqrt{(x-c)^2}.$$