Elegant way to see that $(A\cos(t+\alpha), B\sin(t+\beta))$ always defines an ellipse?

Question

Let $A,B>0$ and $\alpha,\beta\in\mathbb{R}$ and consider the curve $$\gamma(t)=(A\cos(t+\alpha), B\sin(t+\beta)).$$ Is there a nice way to see that it is always an ellipse centered at the origin? It is clear when $\alpha = \beta$ but I cannot see how to handle the general case without getting involved in some nasty calculations.

Answer 1

Adopting @Semiclassical's expression, we know $$y=B\sin(t+\delta)=B(\sin t\cos\delta+\sin\delta\cos t),$$ and so, regrouping a little bit we find that $$\gamma(t)=(A,B\sin\delta)\cos t+(0,B\cos\delta)\sin t= \begin{pmatrix} A & 0\\ B\sin\delta & B\cos\delta \end{pmatrix} \begin{pmatrix} \cos t\\ \sin t \end{pmatrix}$$ which is an affine transformation of the unit circle $\mathbf r(t)=(\cos t,\sin t)$ (by definition an ellipse).

It is centered at the origin because the transformation does not include any translation. Otherwise you can argue that the modules of $x$ and $y$ tend to $0$ as $A$ and $B$ tend to $0$ too, meaning, yet again, that the ellipse is centered at the origin.

Answer 2

First note that the origin is a center of symmetry, since $\gamma(t+\pi)=-\gamma(t)$.

Assume now wlog $\beta=0$. $$\exists t\in\Bbb R,\quad x=A\cos(t+\alpha),\quad y=B\sin t$$ $$\iff\exists t\in\Bbb R,\quad\sin t=\frac yB,\quad x=A\left(\cos t\cos\alpha-\frac yB\sin\alpha\right)$$ $$\iff\exists t\in\Bbb R,\quad\sin t=\frac yB,\quad\cos\alpha\cos t=\frac xA+\frac yB\sin\alpha$$ hence the curve is:

• if $\cos\alpha=0$: the degenerate ellipse (line segment) of equation $$y\in[-B,B],\quad x=-(\sin\alpha)\frac ABy,$$
• if $\cos\alpha\ne0$: the ellipse of equation$$\frac1{\cos^2\alpha}\left(\frac xA+\frac yB\sin\alpha\right)^2+\left(\frac yB\right)^2=1.$$

Answer 3

You can observe that $\gamma(t)$ can be written as

$$ \gamma(t)=\begin{pmatrix}\cos(\alpha) & -\sin(\alpha) \\ \sin(\beta) & \cos(\beta)\end{pmatrix}\begin{pmatrix} A\cos(t) \\ B\sin(t)\end{pmatrix}=A_{\alpha,\beta}\begin{pmatrix} A\cos(t) \\ B\sin(t)\end{pmatrix}.$$

So $\gamma$ is the image of the usual ellipse of equation

$$ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1$$

with respect to the application $A_{\alpha,\beta}$. Let us study this application. We can observe that

$$A_{\alpha,\beta}\begin{pmatrix} x\\ y\end{pmatrix}=pr_{e_1}\left(rot_\alpha \begin{pmatrix} x\\ y\end{pmatrix}\right)+pr_{e_2}\left(rot_\beta \begin{pmatrix} x\\ y\end{pmatrix}\right)$$

where $pr_{e_i}$ are the projections over the $x$ and $y$ axes respectively. In other words, this application sends each vector to the sum of the projection of its rotation by an angle $\alpha$ over the $x$-axis, and the projection of its rotation by an angle $\beta$ over the $y$-axis.

However, it is not still clear why the image of $A_{\alpha,\beta}$, hence $\gamma$, should be an ellipse. The determinant of this matrix is equal to $\cos(\alpha-\beta)$, that it is different from zero if $\beta\neq \alpha+k\frac{\pi}{2}$.

Let us assume for simplicity that $\beta\neq \alpha+k\frac{\pi}{2}$. Then the application is an isomorphism and the equation of the image of the usual ellipse is then

$$ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1 \implies \begin{pmatrix} x & y\end{pmatrix}\begin{pmatrix} \frac{1}{A^2} & 0 \\ 0 & \frac{1}{B^2}\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix}=1 \implies \begin{pmatrix} x' & y'\end{pmatrix}(A_{\alpha,\beta}^{-1})^t\begin{pmatrix} \frac{1}{A^2} & 0 \\ 0 & \frac{1}{B^2}\end{pmatrix}A_{\alpha,\beta}^{-1}\begin{pmatrix} x'\\ y'\end{pmatrix}=1$$

Thus $\gamma$ is still a conic (this is general fact, the image of a conic by a linear map is still a conic). However, $A_{\alpha,\beta}$ is a homeomorphism, so it is sending the ellipse to something homomorphic to it, and that it is still a conic. Hence the image via $A_{\alpha,\beta}$, namely the curve $\gamma$, is still an ellipse.

Looking at how the map is working, it should be clear that the center of the ellipse is still the origin.

Can you study the remain cases where $\beta= \alpha+k\frac{\pi}{2}$?