Transforming a cartesian equation to a polar one when it has different x and y denominators?

Question

$$\frac{x^2}{9}+\frac{y^2}{16}=1$$

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 $x^2+y^2=r^2$?

Answer 1

$$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}=1}$$ 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 $$r^2(16\cos^2\theta+9\sin^2\theta)=144$$

Answer 2

The choice $x=r\cos\theta$, $y=r\sin\theta$ is right. You can then if you wish give various equivalent expressions, by using trigonometric identities. Replacing $\cos^2\theta$ by $\frac{\cos 2\theta+1}{2}$, and $\sin^2\theta$ by $\frac{1-\cos 2\theta}{2}$ is a popular choice.