How to show $|F_1-v|+|F_2-v|=c$ for an ellipse with foci $F_1, F_2$ I am asked to show the sum of the distances between the two foci on the ellipse and any point, $v$, on the ellipse is independent of one’s choice of $v$. 
Starting with the standard equation of an ellipse: $$\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$$ and a parametrization of it: $$\gamma(t)=\left(p\cos(t), q\sin(t)\right)\space\space \text{for}\space t\in[0,2\pi]$$
The foci of the ellipse are $F_1=(ep,0)$ and $F_2=(-ep,0)$ where $e=\sqrt{1-\frac{q^2}{p^2}}$. My attempt to show the sum is constant is the following: 
$$|v-F_1|=\sqrt{(p\cos(t)-ep)^2+(q\sin(t)-0)^2}=\sqrt{p^2\cos^2(t)-2p^2e\cos(t)+e^2p^2+q^2\sin^2(t)}$$ $$|v-F_2|=\sqrt{(p\cos(t)+ep)^2+(q\sin(t)-0)^2}=\sqrt{p^2\cos^2(t)+2p^2e\cos(t)+e^2p^2+q^2\sin^2(t)}$$
I am stuck here trying to show the sum of these two expressions is independent of $t$. 
I would be very grateful for hints and not full solutions. Thanks in advance! 
 A: Per Blue’s request, I will finish my answer outlined above...
$$|v-F_1|=\sqrt{(p\cos(t)-ep)^2+(q\sin(t)-0)^2}=\sqrt{p^2\cos^2(t)-2p^2e\cos(t)+e^2p^2+q^2\sin^2(t)}$$ $$|v-F_2|=\sqrt{(p\cos(t)+ep)^2+(q\sin(t)-0)^2}=\sqrt{p^2\cos^2(t)+2p^2e\cos(t)+e^2p^2+q^2\sin^2(t)}$$ 
Use the change of variable $e=\sqrt{1-\frac{q^2}{p^2}}\implies q^2=p^2-p^2e^2\implies$ \begin{equation} \label{eq1}
\begin{split}
|v-F_1|& = \sqrt{p^2\cos^2(t)-2p^2e\cos(t)+e^2p^2+(p^2-p^2e^2)\sin^2(t)} \\
 & = \sqrt{p^2\cos^2(t)-2p^2e\cos(t)+e^2p^2+p^2\sin^2(t)-p^2e^2\sin^2(t)}\\
&=\sqrt{p^2-2p^2e\cos(t)+e^2p^2-p^2e^2\sin^2(t)} \\
& =\sqrt{p^2-2p^2e\cos(t)+p^2e^2\cos^2(t)}\\
&=\sqrt{p^2\left(1-2e\cos(t)+e^2\cos^2(t)\right)} \\
&=\sqrt{p^2\left(1-e\cos(t)\right)^2} \\
&=p(1-e\cos(t))
\end{split}
\end{equation}
A similar lengthy calculation shows that: $$|v-F_2|=p(1+e\cos(t))$$
Hence $|v-F_1|+|v-F_2|=p(1-e\cos(t))+p(1+e\cos(t))=2p$
Thus we have shown that the sum of the distances between both foci and a point, $v$, on the ellipse is constant namely $2p$.  
