# How to prove that the axis of an ellipse are perpendicular

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,

• Which definition(s) of an ellipse are you allowed to use? Sep 29 at 13:00
• And how do you define the axes? Sep 29 at 13:05
• To emphasize the comments so far: there are multiple ways to define an ellipse. I can give you a definition of ellipse from which the proof is very short. But since I don't know your definition of an ellipse, you might say that's not what I asked, and hence I am disinclined to attempt an answer. Please clarify your post by giving the definition you use. Sep 29 at 13:05
• How do you construct the ellipse? Is it via conic section, or with a directrix, or algebraically or ... Sep 29 at 16:29
• Kettle of fish? Sep 29 at 18:25

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

• I think there are some missing bits in the proof, you havent proved that the ellipse is symmetrical, so you havent proved that twice OP is the shortest segment trough O. (By not having proved symmetry the segment isn't proven the same length at both sides of O). Also the median formula needs a proof.and you even after that you haven't proven that OP is the shortest Sep 30 at 11:40
• @Whogius Symmetry is trivial and is left to the reader. The proof of median formula can be found online. Sep 30 at 19:29
• Thanks, but even with having proved symmetry , (i think).you still.havent proved that OP is the shortest segment containing the middpoint and a point on the ellipse. But i guess you can prove this by proving that the circle with centre O containing P is tangent to the ellipse. Oct 1 at 1:28
• @Whogius What's wrong with "the minimum length of 𝑟=𝑂𝑃 occurs for 𝑥=𝑎" and following? Oct 1 at 7:40
• @Whogius "proving that the circle with centre $O$ containing $P$ is tangent to the ellipse" seems far harder than the proof in the answer using the fact that $(a-x)^2\ge 0$. Oct 1 at 10:21

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.

• Assumes a coordinate system. That was not a given, the formula is not given , Originally my puzzle for mysefl was about a hyperbole, but tthen later realised for an ellipse it was allready to complicated Sep 30 at 12:29
• Works for hyperbolas as well. I believe it covers parabolas as a limiting case. For establishing a coordinate system, a line through the foci serves as an x axis. Circles centered at either foci passing through the other will intersect at two points. A line through those intersections will give you the perpendicular bisector of the line giving you the yaxis and establishing the origin. A circle centered at the origin of arbitrary radius establishes scaling on either axis. An coordinate system follows from Euclid's postulates. Agree , more interesting without. I pauca's is great. Sep 30 at 17:30

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.

• Sorry, assumes allready that the co-vertex is on the y axis, while that needs to be proved , also a o whole orthogonal coordinate system where is the proof of that? Sep 30 at 11:52
• At my level one has to make a lot of assumptions - for simplicity. Also, aren't all ellipses but that ellipse with center at $(0,0)$, subjected to rigid/isometric transformations. The relationships of the parts of the ellipse should - I suppose - remain *unperturbed. I am, of course, only surmising all this. Sep 30 at 12:29
• I understand, putting the centre at (0,0).and asuming he long axis is the x axis, are in themself is not a problem , but then asuming that the other axis is the y axis , that is was just the thing that needed to be proved Sep 30 at 14:12
• Thanks @Whogius Sep 30 at 18:14