The area of a quadrilateral formed with the focii of the conics $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ is?

Question

The points of the quadrilateral are $(\pm ae_1,0)$ and $(0,\pm be_2)$

The area of half of the quadrilateral (a triangle) is

$$\Delta =\frac 12 (2ae_1)(be_2)$$ $$\Delta =abe_1e_2$$

Also $$e_1=\sqrt {1-\frac{b^2}{a^2}}$$ $$e_2=\sqrt {1+\frac{a^2}{b^2}}$$

Therefore $$\Delta =ab\sqrt {\frac{a^4-b^4}{a^2b^2}}$$ Area of quadrilateral is $$=2\sqrt {a^4-b^4}$$

But the answer is $2(a^2+b^2)$. Where am I going wrong?

Answer 1

The equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ represents an ellipse, while $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ represents an hyperbola with foci on $y$ axis. If $a<b$, then the area of the quadrilateral is $0$ because the foci belongs to $y-$axis. In fact, see this graph:

When $a>b$, we have this graph:

In particular, we know that the coords of $A$ and $B$ are: $$(0,\pm c) \rightarrow (0,\pm\sqrt{a^2+b^2})$$ The coords of $C$ and $D$ are: $$(\pm c_1,0)\rightarrow (\pm\sqrt{a^2-b^2},0)$$ where $c,c_1$ are the focal lenght. Now, the quadrilateral $ABCD$ is a romboid, so the area is: $$A_{ABCD}=4\cdot A_{AOB}=4\cdot\frac{1}{2}\cdot\sqrt{a^2+b^2}\cdot\sqrt{a^2-b^2}=2\cdot\sqrt{a^4-b^4}$$

Note that the given answer is wrong since it's referering to the area of the quadrilateral whith vertices the foci of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$. In this case the foci have coords: $$(0,\pm\sqrt{a^2+b^2}) \land (\pm\sqrt{a^2+b^2},0)$$ This tells us that they are the vertex of a square with side lenght: $$V=\sqrt{2}\sqrt{a^2+b^2}$$ And so the area is: $$V^2=(\sqrt{2}\sqrt{a^2+b^2})^2=2(a^2+b^2)$$

Answer 2

The area of the wuadeilater if $a>b$ needs to be $Q=\frac{1}{2} (2ae_1) (2b e_2)$. So the area of the quadrilateral is $$Q=2abe_1e_2=2\sqrt{a^4-b^4}~ if ~~a>b$$. But if $a<b$, all four foci will be collinear so no quadrilateral is formed.