This is the last question on my assignment and I can't figure out how to solve it. Replace the Cartesian equation with an equivalent polar equation: $$ \frac{x^2}{4} + \frac{y^2}{49}=1 $$
I know that $x^2+y^2=r^2$ but when I try to work this out...
$$ \frac{4y^2+49x^2}{196}=1 $$ $$ \frac{4(r\sin\theta)^2+49(r\cos\theta)^2}{196}=1 $$ $$ \frac{4r^2\sin^2\theta+49r^2\cos^2\theta}{196}=1 $$ $$ r^2(4\sin^2\theta+49\cos^2\theta)=196 $$ $$ r^2=\frac{196}{49\cos^2\theta+4\sin^2\theta} $$ $$ r=\sqrt{\frac{196}{49\cos^2\theta+4\sin^2\theta}} $$
But I know this is very wrong.
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$\begingroup$You are totally correct : You have done what is done to find polar form of any equation, i.e.
Assume the polar coordinates of a curve to $\big(r(\theta)\cos \theta,r(\theta)\sin \theta \big)$ and put it into the cartesian equation of the curve, and then solve for $r(\theta)$.
For an ellipse :
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
The polar coordinates of point $\text{P}$ are given by :
$$\text{P} \equiv\Big(r(\theta)\cos \theta,r(\theta)\sin \theta\Big) ; ~\text{where}~r(\theta)=\frac{ab}{\sqrt{a^2 \sin^2 \theta+b^2 \cos^2 \theta}} $$
You can confirm yourself here on Wikipedia.
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