# How do I find the area of an inscribed triangle with a vertex at the intersection of an ellipse and a hyperbola with the same foci?

I'm having some trouble with a problem that asks for the area of a triangle with a vertex at the intersection of an ellipse and a hyperbola. Here it is:

An ellipse and a hyperbola have the same foci - $F_1$ and $F_2$. These curves cross at 4 points - let $P$ be one of the points. These curves also intersect the line $F_1F_2$ at 4 points labelled $Q$, $R$, $S$ and $T$ in that order. If $RS = 20$, $ST = 14$ and $∆PF_1F_2$ is isosceles, compute the area of $∆PF_1F_2$.

Here is what I have so far. Let $a_e, b_e$ be the parameters for the ellipse, $a_h, b_h$ be the parameters for the hyperbola, and $\hat x, \hat y$ be the coordinates of $P$. Assume both curves are centered at the origin and $F_1$ has a negative x-coordinate. I deduced that

$$$$\begin{split} \begin{gathered} a_e = \frac {20 + 2 * 14} 2 = 24\\ a_h = \frac {20} 2 = 10\\ 2c \neq PF_1 \rightarrow 2c = PF_2\\ 2c = a_h - \left( \frac c {a_h} \right) \hat x = 10 - \left( \frac c {10} \right) \hat x \end{gathered} \end{split}$$$$

However, I'm not sure where to go from here. I had a thought of trying to solve for $\hat x$ in terms of $c$, then plugging it into the equation for the ellipse and trying to solve for $\hat y$, then doing $\frac 1 2 * 2c * \hat y$ would give me the area of the triangle purely in terms of $c$. But then there would still be a $b_e$ floating around, and I don't know what $c$ is... any advice on solving this?

• Find the focal distance and the length of the third side of the triangle and use Heron’s formula. – amd Mar 12 '17 at 19:26
• To continue the way you’re going, you’ll need to find a value for the focal distance $f$, which will determine both semi-minor lengths $b_e$ and $b_h$, such that the distance from $P$ to the farther focus is equal to $2f$. The triangle’s area will then be $|yf|$, where $y$ is the abcissa of $P$. That seems to me like way more work than the other method I’ve suggested, though. – amd Mar 12 '17 at 20:16

We know that the apex of the triangle must be one of the foci. If it were $P$, then an intersection of the two conics would lie on the ellipse’s minor axis, which is not possible with confocal conics with the given vertices. Wlog take the apex to be at $F_1$.
By Heron’s formula, the area of the triangle is $$A=\sqrt{s\,(s-|F_1P|)\,(s-|F_2P|)\,(s-|F_1F_2|)}$$ where $s$ is the semi-perimeter $(|F_1P|+|F_2P|+|F_1F_2|)/2$. Setting $|F_1F_2|=|F_1P|=2f$, $|F_2P|=2d$ and simplifying gives $A=d\sqrt{4f^2-d^2}$.
From the properties of the two conic sections we know that $$|F_1P|+|F_2P|=2f+2d=|QT|=2a_e \\ |F_1P|-|F_2P|=2f-2d=|RS|=2a_h.$$ Solving these equations for $f$ and $d$ yields \begin{align}f&=\frac12(a_e+a_h)\\d&=\frac12(a_e-a_h).\end{align} Plug in the values that you’ve computed for the semi-axis lengths, and you’re done.