Ellipse and hyperbola have the same foci I am trying to solve the following problem:

An ellipse and a hyperbola have the same foci, $A$ and $B$, and intersect at four points. The ellipse has major axis $50$, and minor axis $40$. The hyperbola has conjugate axis of length $20$. Let $P$ be a point on both the hyperbola and ellipse. What is $PA \times PB$?

So I say the center of the ellipse is at $(0,0)$ and the equation of the ellipse is $$\frac{x^2}{25^2}+\frac{y^2}{20^2}=1$$
I calculate that the foci of the ellipse are located at $(15,0)$ and $(-15,0)$.
The general equation of a hyperbola is: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm 1 \quad \cdots \cdots (*)$$ since the length of the conjugate axis is $20$, we can say
$$2a = 20 \implies a = 10$$
Since $a$ is $10$  we search for $b$ with the condition that the hyperbola formed from $(*)$ will have foci at $(15,0)$ and $(-15,0)$.
I get $$b = 5 \sqrt{5}$$
Now plugging into $(*)$ the values I have for $a$ and $b$ and get,
$$\frac{y^2}{125}-\frac{x^2}{100}=-1$$
for the equation of the hyperbola. Now we need one intersection point and for that I used Mathematica and get $$P = \left ( \frac{50}{3}, \frac{20 \sqrt{5}}{3} \right )$$
The whole line of reasoning leads to the following diagram:

With $A$ and $B$ the foci and $P$ one of the points of intersection. I used the distance formula to get the length of $PA$ and $PB$ and got $15$ and $35$ as seen in the diagram. $$15 \times 35 = 525$$
Of course, this is not the answer given, which is $500$. Where did I go wrong?
Thanks to all for their nice solutions.
 A: The $125$ and the $100$ in your hyperbola equation should have been swapped:
$$\frac{x^2}{125}-\frac{y^2}{100}=1$$
This gives $P$ as $\left( \dfrac{25\sqrt5}3,\dfrac{40}{3} \right)$ and $PA \times PB \,$ yields the given answer of $500$.
A: Here's a geometric solution.
For any point $P$ on the ellipse, the sum of distances $PA+PB$ to the foci $A$ and $B$ is constant and equals the length of the major axis. So $PA+PB=50$.
Also, for an ellipse the distance $f$ from the center of the ellipse to one of the foci satisfies the equation $f^2+b^2=a^2$, where $2a$ and $2b$ are the lengths of the major and minor axes respectively. So $f^2=25^2-20^2=225$ and therefore $f=15$.
For a hyperbola the distance $f_H$ from the center to one of the foci satisfies the equation $f_H^2=a^2+b^2$, where now $2a$ and $2b$ stand for the lengths of the transverse and conjugate axes respectively. Since the ellipse and hyperbola share the same foci, we know $f_H=f=15$. But we are given that $2b=20$, so $a^2=15^2-10^2=125$ and therefore $a=5\sqrt 5$.
Finally, for a hyperbola the difference of distances $PA-PB$ is constant and equals the length of the transverse axis, i.e., $PA-PB=10\sqrt 5$.
We now have two equations in two unknowns: $PA+PB=50$, and $PA-PB=10\sqrt 5$. Solve these and obtain $PA\cdot PB=500$.
A: Using elliptic coordinates:
\begin{align*}
  z     &= c\cosh (\alpha+\beta i) \\
  (x,y) &= (c\cosh \alpha \cos \beta, c\sinh \alpha \sin \beta)
\end{align*}
Ellipse:
$$\frac{x^2}{c\cosh^2 \alpha}+\frac{y^2}{c\sinh^2 \alpha}=1$$

In this case, $$\
\begin{array}{rcccl}
  \text{major axis} &=& 2c\cosh \alpha &=& 50 \\
  \text{minor axis} &=& 2c\sinh \alpha &=& 40
\end{array}$$

Hyperbola:
$$\frac{x^2}{c\cos^2 \beta}-\frac{y^2}{c\sin^2 \beta}=1$$

In this case, $$\
\begin{array}{rcccl}
  \text{transverse axis} &=& 2c\cos \beta &=& 2\sqrt{15^2-10^2} \\
  \color{red}{\text{conjugate axis}}  &=& 2c\sin \beta &=& 20
\end{array}$$

Now,
\begin{align*}
  |z+c|+|z-c| &= 2c\cosh \alpha \\
  |z+c|-|z-c| &= \pm 2c\cos \beta \\
  4|z+c||z-c| &= (|z+c|+|z-c|)^2-(|z+c|-|z-c|)^2 \\
              &= 4c^2(\cosh^2 \alpha-\cos^2 \beta) \\
   |z+c||z-c| &= c^2(\cosh^2 \alpha-\cos^2 \beta) \\
              &= c^2(\sinh^2 \alpha+\sin^2 \beta) \\
              &= \left( \frac{40}{2} \right)^2+
                 \left( \frac{20}{2} \right)^2 \\
              &= 500
\end{align*}
