# Lemniscate - two parametrizations

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.

• 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}. Mar 31, 2020 at 23:36
• Fix your algebra at the last step. Apr 1, 2020 at 1:06
• Thanks, hopefully fixed. Apr 1, 2020 at 9:29

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 $$\bbox[5px,border:2px solid #C0A000]{r^2=2\cos(2\theta)}\tag3$$

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 $$\bbox[5px,border:2px solid #C0A000]{(x,y)=\left(\frac{\sqrt2\cos(t)}{\sin^2(t)+1},\frac{\sqrt2\sin(t)\cos(t)}{\sin^2(t)+1}\right)}\tag6$$

• 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. Apr 2, 2020 at 10:36
• 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.
– robjohn
Apr 2, 2020 at 13:10
• Ah okay, perfect, that makes it clear then. Thank a lot. Apr 2, 2020 at 13:12