Convert $^2+6−9=0 $to polar.

Question

I need to convert this rectangular equation to polar. I already took a look at some solutions but I have a problem understanding this part:

how was the square root simplified to three?

Answer 1

Let's try the real computation: with $x=r\cos\phi$ and $y=r\sin\phi$, you get $$ r^2\cos^2\phi+6r\sin\phi-9=0 $$ so that $$ r=\frac{-6\sin\phi\pm\sqrt{36\sin^2\phi+36\cos^2\phi}}{2\cos^2\phi} $$ Can you spot the error in the image?

The final result is correct, but there is $\sin^2\phi$ instead of $\cos^2\phi$ under the square root.

Answer 2

Converting $ x $ and $ y $ into polar coordinates: $$ x = r\cos(\phi)$$ and $$ y = r\sin(\phi)$$

Making the above substitutions: $$ x^{2} + 6y - 9 = r^{2}\cos^{2}(\phi) + 6r\sin(\phi) - 9 = 0 $$

To solve for $ r $ using the quadratic formula, let $ a = \cos^{2}(\phi) $, $ b = 6\sin(\phi) $, and $ c = -9 $. As a result,

$$ r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$$$= \frac{-6\sin(\phi) \pm \sqrt{36\sin^{2}(\phi) - 4(\cos^{2}(\phi))(-9)}}{2(\cos^{2}(\phi))}$$$$= \frac{-6\sin(\phi) \pm \sqrt{36\sin^{2}(\phi) +36\cos^{2}(\phi)}}{2\cos^{2}(\phi)}$$

The second term inside the radical should be $36\cos^{2}{\phi}$, not $36\sin^{2}{\phi}$. Thus the radical should be: $$ \sqrt{36\sin^{2}{\phi} + 36\cos^{2}{\phi}} $$