Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation

Question

This is how I tried so far...

$r^2= 9 \cos 2 ( \theta)$

$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and $r^2= x^2 + y^2$

so, it will become

$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$

Answer 1

multiply both side by $r^2$ and replace $r.cos\theta=x$ and $r.sin\theta=y$ and $r^2=x^2+y^2$ as $cos^2\theta-sin^2\theta=(cos\theta+sin\theta)(cos\theta-sin\theta)$

Answer 2

Hint Polar and Cartesian coordinates are related by the equations $$x = r \cos \theta, \quad y = r \sin \theta,$$ so a Pythagorean trigonometric identity gives $$x^2 + y^2 = (r \cos \theta)^2 + (r \sin \theta)^2 = r^2 (\cos^2 \theta + \sin^2 \theta) = r^2 .$$ Now, simply rearrange and substitute.

Answer 3

Starting with $$r^2=9\cos2\theta$$ Applying the double angle identity $$r^2=9\cos^2\theta-9\sin^2\theta$$ Now multiply both sides by $r^2$, apply $x=r\cos\theta,y=r\sin\theta$ and $x^2+y^2=r^2$ to obtain $$(x^2+y^2)^2=9x^2-9y^2$$ Which, when expanded gives $${x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-9\,{x}^{2}+9\,{y}^{2}=0$$