Respuesta :
[tex]\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}[/tex]
[tex]\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}}[/tex]
[tex]\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)[/tex]
[tex]\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}}[/tex]
[tex]\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)[/tex]
Answer:
r^2=6rsin(theta)+8
Step-by-step explanation:
I took the quiz <3