I have a Lemniscate defined as $$ \|{F_{1}-Z}||^{2} ||F_{2}-Z||^{2}=1 $$ for $ F_1 = (-1,0), F_2=(1,0)$ in $\mathbb{R}^2$

And I should find the parametric equations using these two parametrizations

a) $x=r \cos(\phi)$, $y=r\sin(\phi)$

b) $x=x, y=x \sin(t)$

I understand that the result should be (right?) $$(x,y)=\begin{pmatrix}\frac{a \cos(\phi)}{1+\sin^2(\phi)},&\frac{a\sin(\phi)\cos(\phi)}{1+\sin^2(\phi)} \end{pmatrix} $$

So far I have managed to start from the definition $$\begin{Vmatrix} {-1-x}\\{-y} \end{Vmatrix}^2 \begin{Vmatrix} {1-x}\\{-y} \end{Vmatrix}^2=1$$ $$(1+2x+x^2+y^2)(1-2x+x^2+y^2)=1$$ $$(x^2+y^2)^2-2(x^2-y^2)=0$$ then pluged in the first parametrization and got $$r^2=2(\cos^2(\phi)-\sin^2(\phi))=2\cos(2\phi)$$

And dont really know what to do next, to get the form that is written above. Didnt really try much with the second parametrization so far.

  • $\begingroup$ You can get the proper font and spacing for $\cos$ and $\sin$ using \cos and \sin. For operators that don't have a command of their own, you can use \operatorname{name}. $\endgroup$
    – joriki
    Mar 31, 2020 at 23:36
  • $\begingroup$ Fix your algebra at the last step. $\endgroup$ Apr 1, 2020 at 1:06
  • $\begingroup$ Thanks, hopefully fixed. $\endgroup$ Apr 1, 2020 at 9:29

1 Answer 1


A bit of algebra gives $$ \left(y^2+x^2+1\right)^2=4x^2+1\tag1 $$

Applying $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have $$ \left(r^2+1\right)^2=4r^2\cos^2(\theta)+1\tag2 $$ which has the solutions $r=0$ and $$ \bbox[5px,border:2px solid #C0A000]{r^2=2\cos(2\theta)}\tag3 $$ enter image description here

Setting $y=x\sin(t)$ in $(1)$ gives $$ \left(x^2\sin^2(t)+x^2+1\right)^2=4x^2+1\tag4 $$ which simplifies to $$ x^4\left(\sin^2(t)+1\right)^2=2x^2\cos^2(t)\tag5 $$ and has the solutions $(x,y)=(0,0)$ and $$ \bbox[5px,border:2px solid #C0A000]{(x,y)=\left(\frac{\sqrt2\cos(t)}{\sin^2(t)+1},\frac{\sqrt2\sin(t)\cos(t)}{\sin^2(t)+1}\right)}\tag6 $$ enter image description here

  • $\begingroup$ And can you somehow take r from (3) plug it back into the polar coordinates and get (6)? Or are these two parametrizations equivalent? Thanks. $\endgroup$ Apr 2, 2020 at 10:36
  • $\begingroup$ The parameterizations plot the same curves, but the point corresponding to a given $\theta$ is not the same as the point corresponding to $t=\theta$. What is true, is that $\sin(t)=\tan(\theta)$ produce the same point on the curve. $\endgroup$
    – robjohn
    Apr 2, 2020 at 13:10
  • $\begingroup$ Ah okay, perfect, that makes it clear then. Thank a lot. $\endgroup$ Apr 2, 2020 at 13:12

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