Question

Find a cartesian equation for r=2cos(theta) + 3sen(theta).. please.

Answer 1

use r^2=x^2+y^2

Answer 2

we know x=rcos(theta) and y = rsin(theta)

and x^2 + y^2 = r^2

so x^2 + y^2 = (2cos(theta) + 3sen(theta))^2

x^2 + y^2 = 4cos^2 (theta) + 12sin(theta)cos(theta) + 9sin^2(theta)

x^2 + y^2 = 4x^2/r^2 + 12 * x * y / r^2 + 9y^2/r^2

(x^2 + y^2)^2 = 4x^2 + 12xy + 9y^2

x^4 + 2x^2y^2 + y^4 = 4x^2 + 12xy + 9y^2

x^4 + y^4 - 3x^2 - 8y^2 + 2x^2y^2 - 12xy = 0

Answer 3

not too sure about this, was always bad at polar stuff

Answer 4

is it correct? r(2costeta) + 3(sinteta) ; r2 = 2rcos + 3rsinteta; x2+y2= 2x + 3y; x2-2x + y2 - 3y =0 ¿?