Answer
$y^2 + 6x - 9 = 0$
Work Step by Step
$r = \frac{3}{1 + cos\theta}$
$r(1 + cos\theta) = 3$
$r +rcos\theta = 3$
$\pm \sqrt{x^2 + y^2} +x = 3$ (since $r^2 = x^2 + y^2$ and $x = rcos\theta$)
$\pm \sqrt{x^2 + y^2} = 3 - x$
$(\pm \sqrt{x^2 + y^2})^2 = (3 - x)^2$ (squaring both sides)
$x^2 + y^2 = 9 - 6x +x^2$
$y^2 + 6x - 9 = 0$
The equivalent equation in rectangular coordinates is $y^2 + 6x - 9 = 0$
