I've been solving some problems from my Modeling and Optimisation course, and I got stuck in this one:
Given the ellipse's equation $$\frac{(x+c)^2}{a^2}+\frac{y^2}{b^2}=1,$$ where $c>0$. Say we concluded that the expression in polar coordinates of this ellipse where polar coordinates polar axis in the focus $F_1=(0,0)$ is $$\rho = \frac{\frac{b^2}{a}}{1+\frac{c}{a}\cos\phi}.$$ Prove that if we take polar coordinates with polar axis in the other focus, the expression in polar coordinates of the ellipse is $$\rho = \frac{\frac{b^2}{a}}{1-\frac{c}{a}\cos\phi}.$$
The first polar coordinates refer to this variable change: $$x(t)=\rho(t)\cos(\phi(t))$$ $$y(t)=\rho(t)\sin(\phi(t))$$ where $\rho(t)$ equals the distance from $F_1=(0,0)$ to the point of the ellipse and $\phi(t)$ the angle it makes respecting to the $x$-axis. I'm assuming the notation has not been changed in the last expression, to be said, that $\rho$ and $\phi$ in last expression are the same than in the first one (they are values respecting to $F_1$ and not $F_2$) Is this correct? The question doesn't make it clear to me.
I tried using things like, for example, that $\rho+ \rho'$ is constant (where $\rho'$ is the distance from $F_2$), but I got nothing. Naming $\phi'$ the angle for $F_2$, i tried to find a relation between the first variable change and the new one
$$x(t)=\rho'(t)\cos(\phi'(t))$$ $$y(t)=\rho'(t)\sin(\phi'(t))$$
but I did not come to any conclusion more than $\rho' = \frac{\rho\sin\phi}{\sin\phi'}$, and this didn't seem useful at all.
How can I solve this? Any help or hint will be appreciated, thanks in advance.