Convert r^2(3 - 2cos^2 Θ) = 8 + 4r cos Θ to rectangular equation and find center, vertices, and foci?

Question

Equation: r^2(3 - 2cos^2 Θ) = 8 + 4r cos Θ
Θ = Theta

Having a hard time finding examples on how to convert an equation like this into a rectangular equation in my book/notes, and am looking for the steps it would take to solve this problem for my review in case in comes up on an exam. The problem also ask to describe the conic section, which I'm also not sure about, but being able to convert it to a rectangular equation and finding the center, vertices, and foci is what I am really looking for. Thanks in advance for any help given!

Answer 1

r^2[3-2cos^2t]=8+rcost
x=r cost r^2 =x^2+y^2
so 3(x^2+y^2)-2x^2=8+4x
x^2+3y^2-4x-8=0
You can take it from here

Answer 2

I think that: x² + y² = x + y is perfectly correct. If you prefer we can do some manipulation. Try bringing the x and y to the other side, and completing the square: x² - x + y² - y = 0 x² - x + 1/4 + y² - y + 1/4 = 1/2 (x - 1/2)² + (y² - 1/2)² = 1/2 which reveals this polar function as a circle, centre (1/2, 1/2), radius 1/√2. Hope that helps!