Question

Rewrite in Polar Form.... x^(2)+y^(2)-6y-8=0

Answer 1

$\bf x^2+y^2-6y-8=0\qquad \begin{cases} x^2+y^2=r^2\\\\ y=rsin(\theta) \end{cases}\implies r^2-6rsin(\theta)=8 \\\\\\ \textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8 \\\\\\\ [r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby \begin{array}{llll} \textit{so we need a value there to make}\\ \textit{a perfect square trinomial} \end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there} \\\\\\ \textit{now, bear in mind we're just borrowing from zero, 0} \\\\ \textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\ \left. \qquad \right.\uparrow \\ \textit{so-called "completing the square"} \\\\\\\ [r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta) \\\\\\ r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$