Show $\{(r\cos \alpha,r\sin \alpha):r=\frac{6}{1-2\cos \alpha)}\}$ is the image of $\phi(t)=(-4+2\cosh(t),2\sqrt{3}\sinh(t))$

Question

I want to show that $V=\{(r\cos \alpha,r\sin \alpha):r=\frac{6}{1-2\cos \alpha)}\}$ is the image of $\phi(t)=(-4+2\cosh(t),2\sqrt{3}\sinh(t))$, so I can prove it is a manifold. This function was given to me as a hint, so I suspect this is the function whose image is equal to $V$.

I can rewrite $\phi(t)=(-4+e^t+e^{-t},\sqrt{3}e^t-\sqrt{3}e^{-t})$. I also know that $\text{domain } \phi=\mathbb{R}$.

Then I'm stuck. How do I match this to the polar coordinates?

Answer 1

Write $$x=-4+e^t+e^{-t},\,\,y=\sqrt 3 e^t-\sqrt 3e^{-t}.$$ The goal is to eliminate $t$ from this system. Perform elementary operations to have $$e^t=\frac {\sqrt 3(x+4) + y}{2\sqrt 3}$$ and $$e^{-t}=\frac{\sqrt 3(x+4)-y}{2\sqrt 3}.$$

From these, it follows that $$\frac {\sqrt 3(x+4) + y}{2\sqrt 3}=\frac{2\sqrt 3}{\sqrt 3(x+4)-y},$$ which can be further simplified. Finally set $x=r\cos\alpha,\,\, y=r\sin\alpha.$

Answer 2

Clearly $\dfrac{x+4}{2}=\cosh t$ and $\dfrac{y}{2\sqrt{3}}=\sinh t$ so by identity $\cosh^2t-\sinh^2t=1$ we have $$\left(\dfrac{x+4}{2}\right)^2-\left(\dfrac{y}{2\sqrt{3}}\right)^2=1$$ this hyperbola with $a=2$ and $b=2\sqrt{3}$ has the ellipticity $$e=\dfrac{c}{a}=\dfrac{\sqrt{a^2+b^2}}{a}=2>1$$ and parameter $$p=a\left(e-\dfrac{1}{e}\right)=3$$ then $$r=\dfrac{ep}{1-e\cos\theta}=\dfrac{6}{1-2\cos\theta}$$