Analytic Geometry Polar Coordinates

Question

Convert $r=6\cdot\sin\theta$ to a rectangular equation.

Okay, so I've tried multiplying $\sec\theta$ to both sides so that I would have $r\cdot\sec\theta = 6x/y$, and I've tried many other things, but they don't seem to work. How can I convert this?

Answer 1

hint: we have $$\sin(\theta)=\frac{y}{r}$$ plugging this in your formula we have $$r^2=6y$$ or $$x^2+y^2=6y$$

Answer 2

We have $\frac{r}{6}=\sin(\theta)$, in polar we have $y=r\sin(\theta)$, so our equation $r=6\sin(\theta)$ becomes $$r=\frac{6y}{r}.$$ Then we collect $r$ to both sides and we have $$6y=r^{2}=x^{2}+y^{2}$$ and thus we have $$x^{2}+y^{2}-6y=0.$$ Now we complete the square and we have $$(y-3)^{2}+x^{2}=9.$$

Answer 3

$y=r\sin \theta$ and $x=r\cos \theta$

$\frac{dy}{d \theta} = r \cos \theta=x$ and $\frac{dx}{d\theta}=-r\sin \theta=-y$

$\frac{dy}{dx} = - \frac{x}{y}$

$\int ydy=-\int xdx$

$y^2 + x^2 = k = r^2$

$r=6\sin\theta$ and $y=r\sin\theta$

$r^2 = 6\sin\theta \times 6\sin\theta$

$r^2 =6r\sin\theta$ $r^2=6y$

$x^2+y^2=6y$