Name this 2-dim Parametric Curve

Question

is there a name for the following parametric curve? Thanks.

I encountered it when playing around with the tangent lines and normal lines of an ellipse. The parameter $\theta$ is the same parametric angle for the related ellipse that's $\{x = A\cos\theta,\; y = B\sin\theta \}$.

$$ x = \frac{A^2-B^2}{A^2\cos^2\theta + B^2\sin^2\theta }\cdot A\cos\theta \\ y = \frac{A^2-B^2}{A^2\cos^2\theta + B^2\sin^2\theta }\cdot B\sin\theta \\ $$ for any $A>0,\; B>0$

It looks really familiar, doesn't it? The plot looks like a goggle (click to see image) and approaches two separate circles when $A \ll B$. I don't have much clue as to how to make this into polar coordinates.

The Mathematica code for the image is as below

A = 2; B = 5;
ParametricPlot[{
  {  (* ellipse  *)
   A*Cos[u], 
   B*Sin[u]  },
  (A^2 - B^2)/(A^2*Cos[u]^2  +  B^2*Sin[u]^2) { A*Cos[u], B*Sin[u]
    } (* the parametric curve in question *)
  },
 {u, 0, 2 \[Pi]},
 PlotLabel -> "A = " <> ToString[A] <> ", B = " <> ToString[B]
 ]

.

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======== Update =========

In response to Blue's comment, here is how this parametric curve arises:

Take the normal line of a given point on the ellipse $\vec{r_0} \equiv \{x_0 = A\cos\theta,\; y_0 = B\sin\theta \}$: $$ y = y_0 + \frac{ A^2 y_0 }{B^2 x_0} (x - x_0) \qquad \text{or} \qquad B^2 x_0 \cdot y = (B^2 - A^2) x_0 y_0 + A^2 y_0 \cdot x $$

Consider the intercepts of this normal line with the axes: $$ y \to 0 \implies x_n \equiv \frac{A^2 - B^2}{A^2} x_0 \\ x \to 0 \implies y_n \equiv \frac{B^2 - A^2}{B^2} y_0 $$ Note that $x_n$ and $x_0$ are of opposite sign when $A < B$, while $y_n$ and $y_0$ are of opposite sign when $A > B$.

Now, an invariance can be formed by these intercepts: $$ \vec{r_1} \equiv \{ x_n, -y_n \} \implies \vec{r_0} \cdot \vec{r_1} = A^2 - B^2 \\ \vec{r_2} \equiv \{ y_n, x_n \} \implies \vec{r_0} \times \vec{r_2} = A^2 - B^2 $$ The inner product and the outer product (cross) can be considered equivalent forms of the same thing, but geometrically the outer product gives the signed triangle area whereas the inner product is more difficult to show in a plot.

The parametric curve in question is an attempt to visualize $\vec{r_0} \cdot \vec{r_1}$, for example as seen in this diagram for $A=3,\; B=1$:

• the red dot marks $\vec{r_1} = \{ x_n, -y_n \}$,

• the dim red dot marks $\{ x_n, y_n \}$ as a reference,

• the green bigger ellipse is the locus of $\vec{r_1}$ and also of the reference point,

• the blue dot is the point on the ellipse $\vec{r_o}$,

• the ray coming out of the blue dot is the normal line, where the intercepts $(x_N, 0)$ and $(0, y_N)$ are also marked,

• the ray coming out of the red dot shows the projection (for inner product) of $\vec{r_1}$ onto the dim blue thick ray for $\vec{r_0}$,

and finally,

• the yellow-ish curve is the locus of the parametric curve in question, being the intersection of the aforementioned gray thin ray projected onto the thick blue ray.

Note that the yellow curve is not really demonstrating the invariance $A^2 - B^2$, which is just a factor in the parametric equation. I'm still working on visually showing the invariance as an inner product, and this goggle-like curve is so far just a delightful by-product along the way.

Thank you for your time.

Answer 1

Using complex number as coordinates, we have

\begin{align*} z &= \frac{(a^2-b^2)(a\cos \theta+bi\sin \theta)}{|a\cos \theta+bi\sin \theta|^2}\\ |z| &= \frac{|a^2-b^2||a\cos \theta+bi\sin \theta|}{|a\cos \theta+bi\sin \theta|^2} \\ |z| &= \frac{|a^2-b^2|}{|a\cos \theta+bi\sin \theta|} \\ \arg z &= \tan^{-1} \left( \frac{b}{a} \tan \theta \right) \end{align*}

It's an inversion of the ellipse $a\cos \theta+bi\sin \theta$ about the circle $|z|^2=|a^2-b^2|$.

Note Briefly:

For parametrization in polar coordinates, the ellipse should be

$$r = \frac{a b}{\sqrt{a^2\sin^2 \phi+b^2\cos^2 \phi}}$$

and its inversion should be

$$r = \frac{|a^2-b^2|\sqrt{a^2\sin^2 \phi+b^2\cos^2 \phi}}{ab}$$