# Find a polar equation for the curve of the given Cartesian equations: $y=x$, $4y^2 = x$ and $xy=4$

I am asked to find a polar equation for the curve of the given Cartesian equations: $y=x$, $4y^2 = x$ and $xy=4$.

What I got here so far is

$$y = x\\ r \sin(\theta) = r \cos(\theta)\\ \boxed{\tan(\theta) = 1}$$

$$4y^2=x\\ 4r^2 \sin^2(\theta) = r \cos(\theta)\\ \boxed{r = \frac{\cot(\theta)\cdot \csc(\theta)}{4}}$$

$$xy = 4\\ r^2 \sin(\theta) \cos(\theta) = 4\\ \boxed{r^2 = \frac{8}{\sin(2 \theta)}}$$

Am I on the right path?

• $\tan\theta = 1 \iff \theta = \pi/4 \lor \theta = 5\pi/4$ in $[0,2\pi[$. – Henricus V. Mar 27 '16 at 4:41

Yes, all correct. The first equation has no $r$, only $\theta$ since it is a straight line through origin at $45^0.$