Question #baaae

1 Answer
Aug 3, 2017

(x^2-3x+y^2)^2=9(x^2+y^2).

Explanation:

r=3(1+cos(theta))

In general,

x=rcos(theta), y=rsin(theta).
Squaring and adding gives r^2=x^2+y^2.

You have
r=3(1+cos(theta)).

Then, x=rcos(theta), giving cos(theta)=x/r.
r=3(1+x/r),
r^2=3(r+x),
r^2-3x=3r.

Squaring gives,
(r^2-3x)^2=9r^2.

Replacing r^2 by x^2+y^2 gives a final relation between x and y,
(x^2-3x+y^2)^2=9(x^2+y^2).