Question

If $r>0,-\pi \le \theta \pi$ and $r,\theta$ satisfy $r\sin \theta =3$ and $r=4(1+\sin \theta )$ then find the possible solutions of the pair $(r,\theta )$

Answer 1

Here $r=4(1+\sin \theta )$ and $r\sin \theta =3$

Eliminating $\theta$ from above equations,

$r=4(1+\frac{3}{r})$

$\Rightarrow r^2-4r-12=0$

$\Rightarrow r^2-6r+2r-12=0$

$\Rightarrow r(r-6)+2(r-6)=0$

$\Rightarrow (r-6)(r+2)=0$

$\Rightarrow r=6,-2$

$\Rightarrow r\sin \theta =3$

$\sin \theta =\frac{1}{2}$ and $\sin \theta =\frac{-3}{2}$(neglecting $\sin \theta =\frac{-3}{2}$)

$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$

$(r,\theta )=(6,\frac{\pi }{6})$ and $(6,\frac{5\pi }{6})$ are the required parts.