Simplifying $\frac{p^2q^2(1-\epsilon^2\cos^2t)}{p^2\cos^2t+q^2\sin^2t}$, where $\epsilon=\sqrt{1-(q/p)^2}$ I am currently trying to show that the product of the distances from the focis of an ellipse to the tangent line at any point of the ellipse is a constant. While I thought that the computation is a straightforward plug-in and be done with it, this turned out to be harder than expected.
I know that the end result should be $q^2$, but I have no idea on how to manipulate the expression
$$\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}$$ where $p > q > 0, \epsilon = \sqrt{1 - (q/p)^2}$.
Edit: I had mixed up $\sin$ and $\cos$ in the denominator, so instead of $p^2\cos^2(t) + q^2\sin^2(t)$ the expression should be $p^2\sin^2(t) + q^2\cos^2(t)$.
 A: Marcos already calculated value of your expression if $t=0$ as $q^4/p^2$. If $t=\pi/2$ expression is equal to $p^2$. If $p \ne q$, these values are different, that is in this case your expression is not a constant. If $p=q$ the expression is equal to $p$ or $q$ for any $t$.
A: Let us use that $\cos^2(t)+\sin^2(t)=1$ and that $\epsilon^2=1-(q/p)^2$:
$$\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=\frac{p^2q^2(\cos^2(t)+ \sin^2(t) - (1-(q/p)^2)\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=$$
$$=\frac{p^2q^2(\cos^2(t)+\sin^2(t) - \cos^2(t)+(q/p)^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=\frac{p^2q^2(\sin^2(t)+(q/p)^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=$$
$$=\frac{1}{p^2}\frac{p^2q^2(p^2\sin^2(t)+q^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=q^2$$
Maybe you can think that it is hard come up with this solution since you have to notice that you have to use $\cos^2(t)+\sin^2(t)=1$.
I'll give you another solution, which is the standard method to show that a function is constant. The method consists on taking the derivative and see that this is equal to $0$ (since the only functions with derivative $0$ are the constant functions):
Let $f(t)=\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}$, taking the derivative you will find that:
$$f'(t)=\frac{2p^2q^2(p^2\epsilon^2 - p^2+q^2)}{(p^2\sin^2(t) + q^2\cos^2(t))^2}$$
(As homework you can check that this is indeed the derivative). Now notice that:
$$p^2\epsilon^2 - p^2+q^2=p^2(1-(q/p)^2)-p^2+q^2=p^2-q^2-p^2+q^2=0$$
Hence $f'(t)=0$ and thus $f(t)=c$ with $c\in\mathbb{R}$ a constant. To find the value of $f$ you can plug any value for $t$, for example $f(0)=q^2$, and then $f(t)=q^2$.
A: I'm not sure where you miscalculated, but the product is indeed constant. The ellipse $x=p\cos t,\,y=q\sin t$ has foci $(\pm p\epsilon,\,0)$ with $\epsilon:=\sqrt{1-q^2/p^2}$. A tangent $y=mx+c$ has distance $\frac{c\pm mp\epsilon}{\sqrt{1+m^{2}}}$ to the $\pm$ focus; these distances' product is $\frac{c^2-m^2p^2\epsilon^2}{1+m^2}$. The tangent at has $m=-\frac{q}{p}\cot t,\,c=q\csc t\implies\frac{c^2-m^2p^2\epsilon^2}{1+m^2}=q^2$.
