Below is an image that illustrates my situation

enter image description here

$h$ and $k$ represent the semi-major and semi-minor axis respectively. The green line $a$ is drawn at an angle of $\theta$. The blue line $c$ is a tangent line touching the ellipse at the intersection of $a$ and the edge of the ellipse. The red line $d$ is a line perpendicular to $c$. And finally the angle $\phi$ is the angle between $d$ and the orange dashed line $e$.

So my question is how can I find $\phi$? Let me know if there's any missing information, or if there are any questions.

Oh, and what if the ellipse wasn't axis-alligned? How much would that change things?

  • $\begingroup$ A tangent to an ellipses is the (external) bisector of the angle formed by the lines from the point of tangency to the foci. (See this answer, for example.) Maybe you can use this to reduce the question to a bit of trigonometry? $\endgroup$
    – user21467
    Nov 22, 2014 at 21:55

2 Answers 2


I haven't really found any quick, nice geometrical interpretation, so we need to do some calculations. It isn't too bad, though. For the sake of simplicity we will first assume that the ellipse is centered in $0$.

For an ellipse we get the following equation (since you are interested in angles, we will use polar coordinates):

$$X=\binom{h\cos \alpha}{k\sin\alpha}$$

At this stage you need to be aware that this gives you a parametrization of the curve, but does not tell you anything about what the angle $\alpha$ actually is, geometrically speaking. It turns out that this angle is not the possibly naturally assumed $\theta$ from your drawing. Luckily, it will turn out that we can get a fairly nice formula for that - if you are interested, you can find references to this under 'Eccentric anomaly', which bugged Johannes Kepler quite a lot, apparently.

So, first we will ignore this, use our parametrisation and get a nice result. Second, we will try to change our result to fit what you want.

First step:

Now, in order to get the tangent for a certain point, we derive this with respect to $\alpha$:

$$\frac{\partial X}{\partial\alpha}=\binom{-h\sin\alpha}{k\cos\alpha}$$

This gives us a directional vector of the tangent at the point $X(\alpha )$, or, to put it differently, of your line $c$.

Now, as $d$ is perpendicular to $c$, we get an equation for $d$ or the form $$ d: \quad -h\sin\alpha x+k\cos\alpha y= z\qquad \Leftrightarrow \\ y=\frac{h\sin\alpha}{k\cos\alpha}x+\overline{z}=\frac{h}{k}\tan\alpha +\overline{z}$$

for some constants $z,\overline{z}$. And, seeing that the angle $\phi$ you want to find is nothing else than the tangent of the slope of $d$, we are finished and have $\tan\phi=\frac{h}{k}\tan\alpha$.

(Note that we assumed that $\cos\alpha\neq 0$, but that case is fairly simple - either you look at it geometrically, or just look at what happens to all those trignometric functions.)

Second Step:

Okay, so now we have expressed our result in terms of $\alpha$. That's, well, fairly nice but, as you pointed out, does not quite solve the provlem. Instead, we will have to take a look at how $\alpha$ is dependent on $\theta$. To do this, we need an equation.

If we look at the point of the ellipse determined by your angle $\theta$, it must be on the line $y=\tan\theta x$.

Similarly, for this point of the ellipse, we obviously have $\binom{x}{y}=\binom{h\cos\alpha}{k\sin\alpha}$.

Put together, this leaves us with the following system of equations:

$$y=x\tan\theta \\ x=h\cos\alpha \\ y=k\sin\alpha \\ $$

Now, we have two expressions for $y$, hence we have $x\tan\theta=k\sin\alpha$, and the second equation gives us $x=h\cos\alpha$, put together this leaves us with

$$h\cos\alpha\tan\theta=k\sin\alpha\qquad \Leftrightarrow \\ \tan\alpha=\frac{h}{k}\tan\theta$$

Notice that, this seems almost custom-tailored to what we want, because in the result of the first part, all we need is $\tan\alpha$, had we obtained some more obscure expression, well, then we might still had a bit more work to do.


Putting those two results together, one now obtains $\tan\phi=\frac{h}{k}\tan\alpha=\frac{h^2}{k^2}\tan\theta$.

Looking at this result, I actually don't find it as intuitively clear as I thought the wrong version to be. However, in a weird way, it does make kind of sense. That the angle scales with the value of $\frac{h}{k}$, certainly does, the square is not something I'd immediately think of. However, if you ever happen to do any kind of coordinate transformations, or path integrals, or anything like that, this is a fairly good example why one has all those weird transformation formulas.

Possibly something pointing in the direction of why the first approach cannot be right is if you simply take $\theta=\frac{pi}{4}$ (meaning $x=y$), and plug it all into your equation. If you substitute $h\cos\alpha = x$ and so on, you get the tangent vector $\binom{-h/k}{k/h}$, whereas - if $\alpha$ were $\theta$, we only get $\binom{-h}{k}$, which kind of does not work out since those two will, generally, not be parallel.

So, if you think of an ellipse as a kind of scaled circle, where we apply the weight $h$ to the cosine and $k$ to the sine, we "get" the actual weight $\frac{h^2}{k^2}$ for the tangent, and the fact that for $h=k$ we have $\phi=\theta$, which would be the relation in any ordinary circle, still holds. This is a bit similar to what happens if you look at your problem in a different coordinate system - that would also be a nice way to solve the question, involving nothing but a couple coordinate transformations, but possibly not feasible depending on your knowledge and what you need the problem for.

Now, if our ellipse is not centered at $0$ but aligned with the axes, nothing changes. Translation does not change angles. If we rotate our ellipse by any fixed angle $\gamma$, well, then you will need to subtract $\gamma$ from any angle you get as a result.

  • $\begingroup$ Provided I didn't make a mistake somewhere in my answer, and we have $\tan\phi=\frac{h}{k}\tan\theta$, that would be the logical conclusion, yes. $\endgroup$ Nov 23, 2014 at 11:48
  • $\begingroup$ Huh. Well, seeing as my work was rather a quick run-through of what I remember about trigonometry than any in-depth thoughts, I'd need to look it over. Good on you for actually not believing the result without either working through the proof or testing, though. I'll try to find some time to look at it tomorrow or so. $\endgroup$ Nov 23, 2014 at 23:16
  • $\begingroup$ Okay, looked at it and messed up fairly horribly: The mistake lies within the parametrisation, while with the transformation $\binom{x}{y}\to\binom{a\cos \theta}{b\sin\theta}$ we do still get the whole ellipse, $\theta$ does not directly correspond to the angle $\theta$ in your graphic, meaning we do not run through the ellipse at a constant speed. Sorry about that, will edit my reply soon. $\endgroup$ Nov 24, 2014 at 21:40
  • $\begingroup$ You use $\alpha$ a lot in your explanation, what does it refer to? $\endgroup$
    – BitNinja
    Nov 25, 2014 at 22:39
  • $\begingroup$ We use a parametrization of the ellipse, namely that every point $(x,y)$ can be written as $(h\cos\alpha,k\sin\alpha)$ for some angle $\alpha$. At that point, we do not yet know what that angle is geometrically - if you're interested in that, google 'eccentric anomaly'. It is not strictly necessary for the proof, though. $\endgroup$ Nov 25, 2014 at 23:35

Due to the nature of tangent, negative angles can be returned. So I've put together this piecewise function that gives $\phi$ no matter what $\theta$ is (As long as it's in the range $(0, 360]$). $$\phi (\theta )=\begin{cases} \arctan { \left( \frac { { h }^{ 2 } }{ { k }^{ 2 } } \tan { \theta } \right) ,\quad if\quad 90\ge \theta \ge 0 } \\ 180+\arctan { \left( \frac { { h }^{ 2 } }{ { k }^{ 2 } } \tan { \theta } \right) } ,\quad if\quad 180\ge \theta >90 \\ 270-\left( 90-\arctan { \left( \frac { { h }^{ 2 } }{ { k }^{ 2 } } \tan { \theta } \right) } \right) ,\quad if\quad 270\ge \theta >180 \\ 360+\arctan { \left( \frac { { h }^{ 2 } }{ { k }^{ 2 } } \tan { \theta } \right) } ,\quad if\quad 360\ge \theta >270 \end{cases}\\ $$

I'm adding this as an answer because it's too long for a comment. This is all put together using Some Math Student's answer.


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