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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?

Aditya
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    Did you copy the question correctly? Is the first conic $\frac{x^2}{a^2}{\color{red}-}\frac{y^2}{b^2}=1$? (Since the final answer doesn't depend on whether $a>b$, I think either the question itself has a typo, or you copied the question incorrectly.) – Batominovski Apr 08 '20 at 06:01
  • There are a lot of questions, like this one, which could be answered in a few minutes by using a free geometry software, like GeoGebra, to make a diagram and compute the area. I'd suggest you to try it. – Intelligenti pauca Apr 08 '20 at 07:14
  • @WETutorialSchool yes I did. I thought the question was wrong too, but I asked it here just to be sure. – Aditya Apr 08 '20 at 07:17
  • @Aretino I know we can use graphing software, but I think it works well only for numericals. – Aditya Apr 08 '20 at 07:20
  • If you suspect the text of an exercise to be wrong, just assign some values to the parameters ($a$ and $b$ in this case) and check with GeoGebra. – Intelligenti pauca Apr 08 '20 at 13:11

2 Answers2

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:

enter image description here

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

enter image description here

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: $$V^2=(\sqrt{2}\sqrt{a^2+b^2})^2=2(a^2+b^2)$$

Matteo
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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.

Z Ahmed
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