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 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 $4(x^2+y^2)=4x^2+\color{red}{4}y^2$.