Invert a Polar Lemniscata into a hyperbola

Question

Hello i want invert a lemniscate but i find few information on internet, i know how to find the inverse of a function in rectangular coordinates functions but i dont know how to do it in polar coordiantes and how inverting a lemniscate become a hyperbola?.

i am using python for plot because is an assigment.
this is the equation:

r = (a**2)(np.cos(2rad))

import numpy as np 
import matplotlib.pyplot as plt 
import math 
  
plt.axes(projection = 'polar') 
a=3
rads = np.arange(0, (2 * np.pi), 0.01) 

for rad in rads: 
    r = (a**2)*(np.cos(2*rad))  
    plt.polar(rad,r,'g.')  
plt.show()

grafica python

Answer 1

The set of MacLaurin polar curves obtained by inversion $ r\to \dfrac{a^2}{r}$

$$ r^n = a^n \cos n \theta, r^n = a^n \sec n \theta $$

In the particular case $n=2$ Lemniscate of Bernoulli and rectangular hyperbola are mutually invertible about radius of inversion circle $r=a=1$

 PolarPlot[ {1,Sqrt[Sec[2 t]],Sqrt[Cos[2 t]]},{t,0,2 Pi},PolarGridLines->Automatic]

An inversion transformation between Cartesian coordinates can be written

$$ (x,y)\to \dfrac{a^2\;(x,y)}{x^2+y^2}$$

For $n=1,3$ and inversions: