Find the polar equation that has the same graph as the given rectangular equation. I have used the method of substituting $r \sin θ$ for $y$ and $r\cos θ$ for $x$.
$x^2=8(2−y)$
$r^2\cos^2(θ)=8(2−r\sin θ)$
$r^2(1−\sin^2θ)=16−8r\sin θ$
$r^2=r^2\sin^2θ−8r\sin θ+16$
Would be taking the perfect square of the right side of this problem take me to the answer?
\begin{equation} x^2=8(2-y) \end{equation} \begin{equation} r^2\cos^2\theta=16-8r\sin\theta \end{equation} \begin{equation} r^2\cos^2\theta+8r\sin\theta-16=0 \end{equation} Applying the quadratic formula and simplifying gives \begin{equation} r=\frac{4}{\sin\theta\pm1} \end{equation} But both \begin{equation} r=\frac{4}{\sin\theta+1} \end{equation} and \begin{equation} r=\frac{4}{\sin\theta-1} \end{equation} give the entire parabola, so either is correct. You do not need both.
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