Given:
[tex]r=\frac{5}{1+3\sin \theta}[/tex]Find: Rectangular equation.
Sol:
[tex]r^2=x^2+y^2[/tex][tex]\begin{gathered} y=r\sin \theta \\ \sin \theta=\frac{y}{r} \end{gathered}[/tex][tex]\begin{gathered} r=\frac{5}{1+3\sin \theta} \\ r=\frac{5}{1+\frac{3y}{r}} \end{gathered}[/tex][tex]\begin{gathered} r=\frac{5r}{r+3y} \\ r+3y=5 \\ r=5-3y \\ r^2=(5-3y)^2 \end{gathered}[/tex]Put the value in rectangular equation:
[tex]\begin{gathered} x^2+y^2=r^2 \\ x^2+y^2=(5-3y)^2 \end{gathered}[/tex]