I am trying to determine analytically what the curve given by the implicit equation $(1)$ would look like:
$$\left(x-\left(3+\frac{y}{3}\right)t\right)^{2}+y^{2}=1 \tag{1}$$
where $t\in\mathbb{R}$ is a positive parameter.
My attempt
I see that for $t=0$ it is a circumference centred at the origin. Indeed, plotting it I see that for $t\neq0$ it represents rotated ellipses.
How could this be deduced analytically from the implicit equation of the curve?
Best Answer
Writing:
$$\left(x-\left(3+\frac{y}{3}\right)t\right)^{2}+y^{2}=1$$
in this way:
$$\left(x-\frac{t}{3}y-3t\right)^{2}+y^{2}=1,$$
you get an ellipse (whatever $t$) as the image of unit circle
$$X^2+Y^2=1$$
by affine change of coordinates:
$$\binom{X}{Y}=\begin{pmatrix}1&-t/3\\0&1\end{pmatrix}\binom{x}{y}+\binom{-3t}{0}$$
This change of coordinates must be written in the inverse way (see alias/alibi transformations here or here) in order to be interpreted as the (affine) transform applied to the unit circle that generates the ellipse with parameter $t$:
$$\binom{x}{y}=\begin{pmatrix}1&t/3\\0&1\end{pmatrix}\binom{X}{Y}+\binom{3t}{0}$$
explaining the combined effect of a distortion of the unit circle (brought by the triangular matrix) followed by an horizontal translation.
Remark: this distorsion is called a shear in the direction of $x$ axis, or sometimes a transvection.