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?

  • 1
    $\begingroup$ $\tan\theta = 1 \iff \theta = \pi/4 \lor \theta = 5\pi/4$ in $[0,2\pi[$. $\endgroup$ – 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 $45^0.$


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