# Which conic section has the polar equation r=a sintheta?

Dec 3, 2014

Remember:

$\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$

and

${x}^{2} + {y}^{2} = {r}^{2}$.

By multiplying by $r$,

$r = a \sin \theta \implies {r}^{2} = a r \sin \theta$

by rewriting in rectangular coordinates,

$\implies {x}^{2} + {y}^{2} = a y \implies {x}^{2} + {y}^{2} - a y = 0$

by adding ${\left(\frac{a}{2}\right)}^{2}$,

$\implies {x}^{2} + {y}^{2} - a y + {\left(\frac{a}{2}\right)}^{2} = {\left(\frac{a}{2}\right)}^{2}$

${x}^{2} + {\left(y - \frac{a}{2}\right)}^{2} = {\left(\frac{a}{2}\right)}^{2}$

Hence, it is a circle with radius $\frac{a}{2}$, centered at $\left(0 , \frac{a}{2}\right)$.

I hope that this was helpful.