# Polar conversions of coordinates and parametric equations

Express the polar coordinates $P\left(6, -\dfrac{\pi}{4} \right)$ in Cartesian coordinates.

$\displaystyle x=r\cos{(\theta)} ,\ y=r\sin{(\theta)} \implies x^2+y^2=r^2 \wedge \theta = \tan^{-1}{\left(\frac{y}{x}\right)}$

So, $x=6\cos{\left(-\dfrac{\pi}{4}\right)}=3\sqrt{3}$ and $y=6\sin{\left(-\dfrac{\pi}{4}\right)}=-3\sqrt{3} \implies \textbf{ANSWER: } P(3\sqrt{3}, -3\sqrt{3})$.

The answer which is calculated here here contradicts mine. Why?

Furthermore, the answer provided here for the following:

Identify the conic represented by $r=\dfrac{3}{2-\cos{(\theta)}}$.

$\displaystyle r=\dfrac{3}{2-\cos{(\theta)}} \implies r(2-\cos{(\theta)})=3 \implies 2r-r\cos{(\theta)}=3 \implies 2r=3+r\cos{(\theta)} \implies 2\sqrt{x^2+y^2}=3+x \implies 4(x^2+y^2)=(3+x)^2 \implies 4x^2+y^2=9+6x+x^2 \implies 3x^2-6x+y^2=9 \implies 3(x^2-2x)-3+y^2=6 \implies 3(x-1)^2+y^2=6 \implies \textbf{ANSWER: } \frac{(x-1)^2}{\sqrt{2}^2}+\frac{y^2}{\sqrt{6}^2}=1$, an ellipse.

is seemingly contradicted as well. What is my error in these conversions?

• For your link to wolfram, try inputting the angle in radians rather than degrees and I think you will see you have the correct answer – BadAtMaths Jul 28 '15 at 11:24
• @BadAtMaths Interesting. It requested it in degrees. What of my other "error"? – alxmke Jul 28 '15 at 11:28
• Check your algebra when you expand $4(x^2 + y^2)$ – BadAtMaths Jul 28 '15 at 11:33
• A slight modification to my first comment after reading @mathlove's comment. *You will nearly have the correct answer – BadAtMaths Jul 28 '15 at 11:36

## 1 Answer

For the first, note that $6\cos\left(-\frac{\pi}{4}\right)=6\times \frac{1}{\sqrt 2}=3\color{red}{\sqrt 2}\approx 4.243$. (BadAtMaths has already pointed 'radian-degree' problem. You should get this.)

For the second, note that $4(x^2+y^2)=4x^2+\color{red}{4}y^2$.

• Is it true that $\displaystyle r=\frac{3}{2-\cos{(\theta)}} \implies r(2-\cos{(\theta)})=3 \implies 2r-r\cos{(\theta)}=3 \implies 2r=3+r\cos{(\theta)} \implies 2\sqrt{x^2+y^2}=3+x \implies 4(x^2+y^2)=(3+x)^2 \implies 4x^2+4y^2=x^2+6x+9 \implies 3x^2-6x-3+4y^2=6 \implies 3(x-1)^2+4y^2=6 \implies (x-2)^2+\frac{4y^2}{3}=2 \implies \frac{(x-2)^2}{\sqrt{2}^2}+\frac{y^2}{\sqrt{\frac{3}{2}}^2}=1$? – alxmke Jul 28 '15 at 21:05
• @Alex: No. Note that $3x^2-6x-3\not=3(x-1)^2$ and that $3x^2-6x\color{red}{+}3=3(x-1)^2$. Also, you wrote $(x-2)^2$ instead of $(x-1)^2$. – mathlove Jul 28 '15 at 21:09
• I see! Thanks. I make so many of these little errors that skew my results, shamefully. – alxmke Jul 28 '15 at 21:11