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.
Best Answer
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 $500$.