Converting from Polar Coordinates to Rectangular - Problem 2


I’m converting polar equations to rectangular form and I also want to, in the end, identify the resulting curve. I’m starting out with r equals 4 over 1 plus sine theta.

Now my instinct here is to multiply the 1 plus sine theta out of the denominator, because it will give me on the left side an r sine theta. That’s something I can work with. So let me do that.

I’m going to multiply both sides by 1 plus sine theta. And I get r times 1 plus sine theta equals just 4. So I have r, plus r sine theta equals 4. Now the r sine theta is y, that’s a small victory here, but I still have this r to deal with. So what I want to do is I’m going to isolate the r. I’ll tell you why in a second. So 4 minus y.

Let’s go back to our formulas here. I know that r² is x² plus y². Whenever I have an isolated r, I can square both sides to get an r² and then that will become an x² plus y². So let me do that. I'll square both sides and here I get 16 minus 8 y plus y², and the r² becomes x² plus y². And here I’ll get a little cancellation which is nice the y²'s, are going to go away. I’m left with x² equals 16 minus 8y.

Let me switch places, I’ll bring the 8y over and 16 minus x² and then I’ll divide both sides by 8. I get y equals 2 minus 1/8 x². This is a parabola opening downward with vertex at 0,2.

Parabola, vertex 0,2 opens down. And so it’s kind of interesting that you can get the equation of a parabola in polar. This is what it looks like. We’ll be dealing with more with the conic sections in polar form later on.

polar coordinates rectangular coordinates polar equations parabola vertex Pythagorean theorem sine