Answer
True
Work Step by Step
We are given the equation in polar coordinates:
$r=\dfrac{2}{2+3\sin\theta}$
Rewrite the equation and substitute $r$ and $\sin\theta$ using the formulas:
$r=\sqrt{x^2+y^2}$
$y=r\sin\theta$
$r(2+3\sin\theta)=2$
$2r+3r\sin\theta=2$
$2r=2-3r\sin\theta$
$(2r)^2=(2-3r\sin\theta)^2$
$4r^2=(2-3r\sin\theta)^2$
$4(x^2+y^2)=(2-3y)^2$
$4x^2+4y^2=4-12y+9y^2$
$4x^2-5y^2+12y-4=0$
Determine the coefficients $A,B,C$ from the general equation:
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
$A=4$
$B=0$
$C=-5$
Compute $B^2-4AC$:
$B^2-4AC=0^2-4(4)(-5)=80>0$
Because $B^2-4AC>0$, the equation represents a hyperbola. Therefore the statement is TRUE.
