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\begin{align} (r\cos\phi)^2 & + 6 r \sin\phi- 9 = 0\\ (r\cos\phi)^2 & = 9 - 6r \sin\phi \end{align}

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Where is the question then? –  awllower Feb 5 '13 at 7:47
    
I am not in class rather stuck on problem –  user1530249 Feb 5 '13 at 8:11

3 Answers 3

Since $r^2\cos^2(\phi)+6r\sin(\phi)-9=0$, $$ \begin{align} r &=\frac{-6\sin(\phi)\pm\sqrt{36\sin^2(\phi)+36\sin^2(\phi)}}{2\cos^2(\phi)}\\ &=\frac{-3\sin(\phi)\pm3}{\cos^2(\phi)}\\ &=\frac3{\sin(\phi)\pm1} \end{align} $$ Note that both '$+$' and '$-$' yield the same curve. '$-$' simply gives the curve $180^\circ$ around with negative the radius. So we can simply pick '$+$': $$ r=\frac3{1+\sin(\phi)} $$

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$$x^2+6y=9$$

As always we choose $x=r \cos \theta$ and $y=r \sin \theta$

Thus $$(r \cos \theta)^2 +6r \sin \theta -9=0$$

With the identity $\cos^2 \theta = 1- \sin^2 \theta$ we find

$$r^2 - (r \sin \theta)^2 + 6r \sin \theta -9=0$$

Which equals $$(r \sin \theta -3)^2=r^2$$

Thus $r=r \sin \theta -3$ or $r= 3-r \sin \theta$ and subsequently we see $$r=\frac{3}{\sin \theta - 1} \lor r=\frac{3}{1+ \sin \theta} $$

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Note that both $r=\frac3{\sin(\theta)-1}$ and $r=\frac3{1+\sin(\theta)}$ represent the same points, but at $\theta$ for one and $\theta+\pi$ for the other. –  robjohn Feb 5 '13 at 16:29

You could also solve for $r$ as a function of $\theta$: write $\cos^2(\theta)$ in terms of $\sin(\theta)$, put all terms on the left, and factor. You should end up with $$ r = \frac{3}{1+\sin(\theta)}$$

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Thanks this is the right answer. Could you explain how you went from r2cos2(θ)+6rsin(θ)=9 to the answer you posted above? –  user1530249 Feb 5 '13 at 8:25
    
@user1530249: we have $r^2\cos^2(t)+6r\sin(t)-9=0$. $\cos^2(t)=1-\sin^2(t)$, so $r^2(1-\sin^2(t))+6r\sin(t)-9=0$, so $r^2-r^2\sin^2(t)+6r\sin(t)-9=0$. This is a quadratic equation and can be solved easily. –  Babak S. Feb 5 '13 at 12:57

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