Transforming a cartesian equation to a polar one when it has different x and y denominators? $$\frac{x^2}{9}+\frac{y^2}{16}=1$$
Needs to be replaced with an equivalent polar equation.  I'm sure the identity I'll have to use will be $$x^2+y^2=r^2$$
though other options are: 
$x=rcos\theta$ and $y=r\sin\theta$.
Originally I tried to simply eliminate the denominator.  Then I tried to take the square root of both sides to get $\frac{x}{3}+\frac{y}{4}=1$,
to use an angle-sum identity with $r\cos\theta*\frac{1}{3}+r\sin\theta*\frac{1}{4}=1$ to reduce it to $r\cos(\theta+something)$ or $r\sin(\theta+something)$.  Neither work.
How should I approach this problem to get it into the form $x^2+y^2=r^2$?

$$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}=1}$$  does not mean $$\frac{x}{3}+\frac{y}{4}=1$$
as $$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}}\ne\frac x3+\frac y4$$
Set $x=r\cos\theta$ and $y=r\sin\theta$ in $16x^2+9y^2=144$ 
to get $$r^2(16\cos^2\theta+9\sin^2\theta)=144$$
