Prove that this is the equation of an ellipse

Question

$$r(\phi)=\frac{c}{1+\epsilon\cos\phi}$$

Prove that this is the equation of an ellipse for $0\le\epsilon<1$ by converting to Cartesian coordinates.

$\displaystyle x=r\cos\phi=\frac{c\cos\phi}{1+\epsilon\cos\phi}$

$\displaystyle y=r\sin\phi=\frac{c\sin\phi}{1+\epsilon\cos\phi}$

How do I eliminate $\phi$? Any suggestions?

Answer 1

$$ \begin{align} r\left(1+\epsilon{x\over r}\right)&=c\\ r+\epsilon x&=c\\ r^2&=(c-\epsilon x)^2\\ x^2+y^2&=(c-\epsilon x)^2\\ (1-\epsilon^2)x^2+2\epsilon cx+y^2&=c^2\\ (1-\epsilon^2)\left(x+{\epsilon c\over 1-\epsilon^2}\right)^2+y^2&={c^2\over 1-\epsilon^2}\\ {(1-\epsilon^2)^2\over c^2}\left(x+{\epsilon c\over 1-\epsilon^2}\right)^2+ {1-\epsilon^2\over c^2}y^2&=1\\ \end{align} $$

Answer 2

You can prove, that major axis lies on X axis, ($\cos$ is even). So the equation of the ellipse is of form:

$$(x-a)^2+b^2y^2=\gamma.$$

Now you insert $x$ and $y$ to this and convert $\sin^2\phi$ to $1-\cos^2\phi$. Now compare expressions with the same powers of $\cos\phi$ in nominator and denominator (with $\gamma$ factor of course) - they must be equal.