Answer
Hyperbola
$\dfrac{(x+2)^2}{1}-\dfrac{y^2}{3}=1$
Work Step by Step
We are given the polar equation:
$r=\dfrac{3}{1-2\cos\theta}$
The equation is in the form:
$r=\dfrac{ep}{1-e\cos\theta}$
Determine $e$ using the denominator:
$e=2$
Determine $p$ using the numerator:
$ep=3$
$2p=3$
$p=\dfrac{3}{2}$
Because $e>1$, the conic is a hyperbola.
Bring the equation to rectangular form:
$r(1-2\cos\theta)=3$
$r-2r\cos\theta=3$
$r=2r\cos\theta+3$
$r^2=(2r\cos\theta+3)^2$
$x^2+y^2=(2x+3)^2$
$x^2+y^2=4x^2+12x+9$
$-3x^2-12x+y^2=9$
$-3(x^2+4x+4)+12+y^2=9$
$-3(x+2)^2+y^2=-3$
$3(x+2)^2-y^2=3$
$\dfrac{(x+2)^2}{1}-\dfrac{y^2}{3}=1$
