Converting a Conic Section Equation from Polar Form to Rectangular Form

In reference to this question -- checking if we did the right thing...



Let's say you ended off with:

\(y^2=3x^2+12x+9\)

To check if you've done it correctly, select a random x value like x=1. Now use this to find out the output:

\(y^2=3\left(1\right)^2+12\left(1\right)+9\)

\(y^2=24\)

\(y=\sqrt{24}\approx 4.8989\)

So, you have a point at (1, 4.8989)

Convert this point to r and theta using the formulas:

\(r=\sqrt{x^2+y^2}\)

\(r=\sqrt{\left(1\right)^2+4.8989}=\sqrt{25}=5\)

\(\theta =\tan ^{-1}\left(\frac{y}{x}\right)\)

\(\theta =\tan ^{-1}\left(\frac{4.8989}{1}\right)=78.4630° \)

Therefore, (1,1) is the same as 5∠78.4630 in polar coordinates.

Now test to see if the left side and right side are equal in the polar equation:

\(r=\frac{6}{2-4\cos \theta }\)

\(r=\frac{6}{2-4\cos \left(78.4630° \right)}=5\)