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