How to prove that the axis of an ellipse are perpendicular

Question

Some visuals are so obvious that you would think proofs are not needed. But then trying to proof them rigouresly is a whole other kettle of fish.

I was stumped by the following puzzle I made for myself: (really it is no homework question)

How do you proof that the axis of an ellipse are perpendicular?

Yes you can see it, it is obvious but seeing in itself is no proof.

Yes you cannot construct a countermodel, but again that is no proof.

I am really stumped with this one, it is so obvious, and easy to see, but a proof?

As definition of an ellipse I want to use: (reused from Wikipedia)

An ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the points is a constant.

As definition of the axis i want to use: (all made up by myself, so maybe incorrect, the second one, defining the minor axis, was a real struggle ;)

The first axis of the ellipse is the line containing the longest segment possible between two points on the ellipse.

The second axis of the ellipse is the line containing the midpoint of the two focal points and the shortest segment possible between two points on the ellipse,

Answer 1

Let $P$ be any point on the ellipse, $F$ and $F'$ its foci, $O$ the midpoint of $FF'$. Set $OP=r$, $FF'=2c$, $FP+F'P=2a$, $FP=x$, $F'P=2a-x$. From the formula for the length of a median we get: $$ r^2=(a-x)^2+a^2-c^2. $$ From that we infer that the minimum length of $r=OP$ occurs for $x=a$. In that case $FP=FP'$, so that $OP\perp FF'$. The segment formed then by $P$ and its reflection about $O$ is the minor axis (according to your definition) and it is perpendicular to $FF'$.

On the other hand, from triangle inequality $FF'\ge |PF'-PF|$ we get: $$ x\ge a-c $$ with equality occurring only when $P$ lies on line $FF'$. Let then $A, A'$ be the intersections of the ellipse with line $FF'$: from the above result it follows that $AA'$ is the longest diameter of the ellipse, and by triangle inequality this is also the longest segment between any two points on the ellipse, hence the major axis (according to your definition). In fact, if $Q$ and $R$ are two points on the ellipse not aligned with $O$, we have: $$ QR<OQ+OR\le2OA=AA'. $$

Answer 2

Suppose you have an arbitrary conic section:

$Ax^2+Bx+Cy^2+Dy+Exy+F=0$

Without loss of generality we can assume $F=1$ since the equation holds if all coefficients are divided by F.

Now suppose $E=0$. Then you can complete the square in $x$ and $y$. Doing so, you find the center of the curve is $(-B/2A, -D/2C)$. This suggests letting $u=x+B/2A$ and $v= y+D/2C$ resulting in a equation of the form $A'u^2+B'v^2+G'=0$. This coordinate change is effectively a translation and translations preserve lines of symmetry. Further, if $(u,v)$ is on the curve then all 4 of $(\pm u, \pm v)$ are on the curve. The x and y axes are therefor lines of symmetry.

If $E\ne 0$, a rotation of the coordinate axes can yield that result.

Let $x=x'\cos\theta -y' \sin\theta$

$y=x'\sin\theta+ y'\cos\theta$ and substitute into the equation for the conic.

$A(x'\cos\theta - y'\sin\theta)^2+B(x'\cos\theta)+C(x'\sin\theta + y'\cos\theta)^2 + D (x'\sin\theta + y'\cos\theta)+E(x'\cos\theta - y'\sin\theta)(x'\sin\theta+y'\cos\theta)+F=0$

Only keep track of the coefficient of $x'y'$ and set it equal to zero. This is the condition for the ellipse to have horizontal and vertical lines of symmetry in the primed coordinate system.

$-2Ax'y'\sin\theta\cos\theta + 2C x'y'\sin\theta\cos\theta + Ex'y'\cos^2\theta-Ex'y'\sin^2\theta=0$

$(C-A)\sin2\theta + E\cos2\theta=0$

$\tan2\theta = \frac{E}{A-C}= \frac{2 \tan \theta}{1-\tan^2\theta}$

$(1-\tan^2 \theta)E=2(A-C)\tan\theta$

$E\tan^2\theta + 2(A-C)\tan \theta -E=0$

$\tan \theta = \frac{-(A-C)\pm \sqrt{(A-C)^2+E^2}}{E}$

The two solutions are the slopes of the lines of symmetry. Note their product is $-1$, the condition for two lines to be perpendicular.

Answer 3

For a horizontal ellipse with the following values:

1. Vertex $(a, 0)$
2. Covertex $(m, n)$
3. Center $(0, 0)$

IF the major and minor axes are perpendicular THEN ...

$a^2 + m^2 + n^2 = (a - m)^2 + (0 - n)^2$
$a^2 + m^2 + n^2 = a^2 - 2mn + m^2 + n^2$
$-2mn = 0$ $mn = 0$

Prove that for this ellipse $mn = 0$ or that $m = 0$ or $n = 0$. We know that $n \ne 0$, otherwise $(m, n) = (a, 0)$. So $m = 0$.

We have to prove $m = 0$

Let $a$ be the major radius and $b$ be the minor radius.

$b^2 = m^2 + n^2$

Substitute that into the equation of an ellipse ...

$\frac{x^2}{a^2} + \frac{y^2}{m^2 + n^2} = 1$
$m^2(m^2 + n^2) + a^2n^2 = a^2(m^2 + n^2)$
$m^2(m^2 + n^2) + a^2n^2 = m^2a^2+ a^2n^2$
$m^2(m^2 + n^2) = m^2a^2$
$m^2 + n^2 = a^2$
So, $m^2b^2 = m^2a^2$
$b \ne a \implies b^2 \ne a^2$
Thus, $m^2 = 0 \implies m = 0$
So the covertex $(m, n)$ is $(0, n)$ i.e. it lies on the $y$ axis. The $y$ axis is orthogonal to the $x$ axis along which lies the major axis of the ellipse. So, the minor axis is also orthogonal to the major axis of an ellipse.