Explicit parametrization for a 3-ellipse? A 4-ellipse?

Question

I searched around but was unable to find anything.

For the usual $2$-ellipse we have the parametrization $x(t) = a\cos(t)$ and $y(t) = b\sin(t)$ for $t\in [0,2\pi]$.

Is there anything similar for the more general $3$-ellipse or $4$-ellipse? If there general case is too difficult is there a parametrization for a constrained version, for example where some of the foci lie, say on the $x$-axis?

Edit: I mean in 2 dimensions, as here https://en.wikipedia.org/wiki/N-ellipse

Edit 2: Assume $u_1 = (-R,0)$ and $u_2 = (R,0)$ and $u_3 = (0, -H)$ are the foci of the $3$-ellipse, for $R,H > 0$, given some $d$ the $3$-ellipse is the set of points given by $$\left\{(x, y) \in \mathbf{R}^{2} : \sum_{i=1}^{3} \sqrt{\left(x-u_{i}\right)^{2}+\left(y-v_{i}\right)^{2}}=d\right\}$$

Is there a parametrization of this curve similar to the $2$-ellipse?

Answer 1

A Parametrization of the 3-Ellipse

Let $$x^2+y^2+z^2=\rho$$ and $$Q_k\equiv (u-u_k)^2+(v-v_k)^2$$ where $u,v$ are our plane variables; then we wish to parametrize $$\sqrt{Q_1}+\sqrt{Q_2}+\sqrt{Q_3}=\rho.$$ Let $$\sqrt{Q_1}=x^2\\ \sqrt{Q_2}=y^2,\\ \sqrt{Q_3}=z^2.$$

Then

$$u^2+v^2-2u_1u-2v_1v+u_1^2+v_1^2=x^4\\u^2+v^2-2u_2u-2v_2v+u_2^2+v_2^2=y^4\\u^2+v^2-2u_3u-2v_3v+u_3^2+v_3^2=z^4.$$

Subtract (2) from (1) and (3) from (2) and (1) from (3) to obtain

$$2(u_2-u_1)u+2(v_2-v_1)v+u_1^2+v_1^2-u_2^2-v_2^2=x^4-y^4,\\2(u_3-u_2)u+2(v_3-v_2)v+u_2^2+v_2^2-u_3^2-v_3^2=y^4-z^4,\\2(u_1-u_3)u+2(v_1-v_3)v+u_3^2+v_3^2-u_1^2-v_1^2=z^4-x^4.$$ From here we relabel $$U_1u+V_1v+W_1=x^4-y^4,\\U_2u+V_2v+W_2=y^4-z^4,\\U_3u+V_3v+W_3=z^4-x^4,$$ and we know that $$\begin{vmatrix}U_1&V_1&-W_1+x^4-y^4\\U_2&V_2&-W_2+y^4-z^4\\U_3&V_3&-W_3+z^4-x^4\end{vmatrix}=0$$ $$\begin{vmatrix}U_1&V_1&x^4-y^4\\U_2&V_2&y^4-z^4\\U_3&V_3&z^4-x^4\end{vmatrix}=\begin{vmatrix}U_1&V_1&W_1\\U_2&V_2&W_2\\U_3&V_3&W_3\end{vmatrix}$$ $$[U_2V_3]^-(x^4-y^4)-[U_1V_3]^-(y^4-z^4)+[U_1V_2]^{-}(z^4-x^4)=[U_1V_2W_3]^-$$ $$\frac{[U_2V_3]^--[U_1V_2]^{-}}{[U_1V_2W_3]^-}x^4-\frac{[U_1V_3]^-+[U_2V_3]^-}{[U_1V_2W_3]^-}y^4+\frac{[U_1V_2]^{-}+[U_1V_3]^-}{[U_1V_2W_3]^-}z^4=1.$$ Let's now parameterize $(x,y,z)$ using $$(x,y,z) = \sqrt{\rho}(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$$ and therefore our second equation restricts one parameter (we'll choose $\theta$):

$$\Delta_x(\sin\theta\cos\phi)^4+\Delta_y(\sin\theta\sin\phi)^4+\Delta_z(\cos\theta)^4=\frac{1}{\rho^2}$$

$$\left[\Delta_x(\cos\phi)^4+\Delta_y(\sin\phi)^4\right]\left(1-(\cos\theta)^2\right)^2+\Delta_z(\cos\theta)^4=\frac{1}{\rho^2}$$

and since the root $x$ of

$$a(1-x)^2+bx^2=c$$ is $$x_{\pm}=\frac{a\pm\sqrt{(a+b)c-ab}}{a+b}$$ we have $$\cos\theta=\sqrt{\frac{\rho(\cos\phi)^2(\Delta_x+\Delta_y(\tan\phi)^4)+\sqrt{(\Delta_x+\Delta_y(\tan\phi)^4+\Delta_z(\sec\phi)^4)-\rho^2(\Delta_x+\Delta_y(\tan\phi)^4)\Delta_z}}{\rho(\cos\phi)^2(\Delta_x+\Delta_y(\tan\phi)^4+\Delta_z(\sec\phi)^4)}}$$ for which $z=\sqrt{\rho}\cos\theta,$ and from that we can derive $\sin\theta=\sqrt{1-\cos^2\theta}.$ Thus our $u,v$ can be obtained from

$$U_1u+V_1v+W_1=x^4-y^4,\\U_2u+V_2v+W_2=y^4-z^4.$$

$$\begin{bmatrix}U_1&V_1\\U_2&V_2\end{bmatrix}\begin{bmatrix}u\\ v\end{bmatrix}=\begin{bmatrix}x^2-y^2-W_1\\ y^2-z^2-W_2\end{bmatrix}$$ or $$\begin{bmatrix}u\\ v\end{bmatrix}=\frac{1}{[U_1V_2]^-}\begin{bmatrix}V_2&-V_1\\-U_2&U_1\end{bmatrix}\begin{bmatrix}x^4-y^4-W_1\\ y^4-z^4-W_2\end{bmatrix}$$ and $(u,v)$ our parametrization is finally given by $$\boxed{\boxed{\left(\frac{V_2x^4-(V_1+V_2)y^4+V_1z^4+[V_1W_2]^-}{[U_1V_2]^-},\frac{-U_2x^4+(U_1+U_2)y^4-U_1z^4+[W_1U_2]^-}{[U_1V_2]^-}\right).}}$$

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Trying to do it more directly in a previous attempt:

This is an attempt at the 3-ellipse:

Let's give it a try. Let $$Q_k=(x-x_k)^2+(y-y_k)^2$$

so that $$\sqrt{Q_1}+\sqrt{Q_2}+\sqrt{Q_3}=\rho.$$

Now let $$\sqrt{Q_3}=\rho \cos^2 \alpha\\ \sqrt{Q_1}+\sqrt{Q_2}=\rho\sin^2\alpha $$ so that

$$Q_3=\rho^2\cos^4\alpha \\ 4Q_1Q_2=\left(\rho^2\sin^4\alpha-Q_1-Q_2\right)^2.$$ For simplicity let $(x_3,y_3)=(0,0),$ and $(x_2,y_2)=(1,0)$ so $$x^2+y^2=\rho^2\cos^4\alpha\equiv g,\\ x^2=g-y^2,\\ \rho^2\sin^4\alpha=(\rho-\sqrt{g})^2\equiv G.$$ Then

$$4((x-1)^2+y^2)((x-x_1)^2+(y-y_1)^2)=\left(\rho^2\sin^4\alpha-Q_1-Q_2\right)^2$$

$$4(g-2x+1)(g-2xx_1+x^2_1-2yy_1+y^2_1)=\left(G-(g-2x+1)-(g-2xx_1+x^2_1-2yy_1+y^2_1)\right)^2;$$ let $x_1^2+y_1^2=Z$ and collecting like terms in $x,y$ we obtain

$$16x_1x^2+16y_1xy-8((1+g)x_1+Z+g)x-8(1+g)y_1y+4(Z+g)(g+1)=\left(G-2g-Z-1+2(1+x_1)x+2y_1y\right)^2$$ which we can relabel as $$ax^2+bxy+cx+dy+e=(u+vx+wy)^2\\ ag+e+dy-ay^2+(c+by)\sqrt{g-y^2}=\left(u+v\sqrt{g-y^2}+wy\right)^2$$ or $$ag+e+dy-ay^2+(c+by)\sqrt{g-y^2}=u^2+v^2g+2uwy+(w^2-v)y^2+2(uv+vwy)\sqrt{g-y^2}$$

$$0=u^2+(v^2-a)g-e+(2uw-d)y+(w^2-v+a)y^2+2(uv-c+(vw-b)y)\sqrt{g-y^2}$$ which can be rephrased as

$$0=A+By+Cy^2+2(D+Ey)\sqrt{g-y^2}$$ $$(A+By+Cy^2)^2-4(D+Ey)^2(g-y^2)=0$$ $$A^2+B^2y^2+C^2y^4+2y(AB+CAy+BCy^2)-4(D+Ey)^2(g-y^2)=0$$ $$A^2+2ABy+(B^2+CA)y^2+2BCy^3+C^2y^4-4(D+Ey)^2(g-y^2)=0$$ $$A^2+2ABy+(B^2+CA)y^2+2BCy^3+C^2y^4+(-4gD^2-8gDEy+4(D^2-gE^2)y^2+8DEy^3+4E^2y^4)=0$$ $$A^2-4gD^2+2(AB-4gDE)y+(B^2+CA+4(D^2-gE^2))y^2\\+2(BC+4DE)y^3+(C^2+4E^2)y^4=0$$ which can again be rephrased as

$$P+Qy+Ry^2+Sy^3+Ty^4=0$$

I'm probably not going to go into detail trying to find out if, for example, $Q,S=0$, or if $T=0$, or if under some transformation I can make enough terms vanish: for now, it's sufficient know that the problem of parametrizing this octic (deg 8) curve needs only the solution to a quartic, and not a septic.

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$$\begin{matrix}\text{Coefficient}&\text{Definition}\\ \\ P&A^2-4gD^2\\ Q&2(AB-4gDE)\\ R&B^2+CA+4(D^2-gE^2)\\ S&2(BC+4DE)\\ T&C^2+4E^2\\ \\ A&u^2+(v^2-a)g-e\\ B&2uw-d\\ C&w^2-v+a\\ D&uv-c\\ E&vw-b \\ \\ a&16x_1 \\ b&16y_1\\ c&-8(Z+g+(1+g)x_1)\\ d&-8(1+g)y_1\\ e&4(Z+g)(g+1)\\ \\ u&G-2g-Z-1\\v&2(1+x_1)\\ w&2y_1\\ \\ Z&x_1^2+y_1^2\\G&(\rho-\sqrt{g})^2\\g& \rho^2\cos^4\alpha\end{matrix}$$

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If I would do it over again, I'd wager the substitution $X=x^2+y^2,Y=x^2-y^2.$