ANSWER
[tex]{r}^{2} = 4 \cos2\theta[/tex]
EXPLANATION
The Cartesian equation is
[tex] {( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )[/tex]
We substitute
[tex]x = r \cos( \theta) [/tex]
[tex]y = r \sin( \theta) [/tex]
and
[tex] {x}^{2} + {y}^{2} = {r}^{2} [/tex]
This implies that
[tex] {( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )[/tex]
Let us evaluate the exponents to get:
[tex] {r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)[/tex]
Factor the RHS to get:
[tex] {r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)[/tex]
Divide through by r²
[tex]{r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)[/tex]
Apply the double angle identity
[tex]\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta) [/tex]
The polar equation then becomes:
[tex]{r}^{2} = 4 \cos2\theta[/tex]
