Respuesta :

SOLUTION

From the question

[tex]x^2+y^2+3y=0[/tex]

This becomes

[tex](x^2+y^2)+3y=0[/tex]

In polar,

[tex]\begin{gathered} x^2+y^2=r^2 \\ \\ \text{and } \\ \\ y=r\sin \theta \end{gathered}[/tex]

So, this becomes

[tex]\begin{gathered} r^2+3r\sin \theta=0 \\ \\ \frac{r^2}{r}=\frac{-3r\sin \theta}{r} \\ \\ r=-3\sin \theta \end{gathered}[/tex]