## Task B-4.6.

An ellipse $x^{2}+4 y^{2}=4$ has a rhombus inscribed in it, with one vertex at point $A(\sqrt{2}, y>0)$. Determine the coordinates of the remaining vertices and calculate the area of the rhombus.
## Solution.

The vertex $A\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right)$, and symmetric to the origin is $\operatorname{vertex} C\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$.

The line $A C$ is a diagonal of the rhombus, passing through the origin and has the equation $y=k x$, where

$$
k=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}
$$

The other diagonal then has the equation $y=-\frac{1}{k} x, \text{i.e.} \, y=-2 x$.

The vertices $B$ and $D$ will be obtained as the intersection of the line $y=-2 x$ and the ellipse.

$$
x^{2}+4(-2 x)^{2}=4 \Longrightarrow 17 x^{2}=4 \Longrightarrow x^{2}=\frac{4}{17}
$$

1 point

from which it follows

$$
x= \pm \frac{2}{\sqrt{17}}, \quad y=-2\left( \pm \frac{2}{\sqrt{17}}\right)=\mp \frac{4}{\sqrt{17}}
$$

that is

$$
B\left(-\frac{2}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right) \quad \text { and } \quad D\left(\frac{2}{\sqrt{17}},-\frac{4}{\sqrt{17}}\right)
$$

The area of the rhombus is $P=\frac{e \cdot f}{2}$, where $e$ and $f$ are the diagonals of the rhombus. Now we have

$$
\begin{aligned}
& e=|\overline{A C}|=\sqrt{(2 \sqrt{2})^{2}+\left(2 \frac{1}{\sqrt{2}}\right)^{2}}=\sqrt{10} \\
& f=|\overline{B D}|=\sqrt{\left(2 \frac{2}{\sqrt{17}}\right)^{2}+\left(2 \frac{4}{\sqrt{17}}\right)^{2}}=\sqrt{\frac{80}{17}} \\
& P=\frac{\sqrt{10} \cdot \sqrt{\frac{80}{17}}}{2}=10 \sqrt{\frac{2}{17}}=\frac{10 \sqrt{34}}{17}
\end{aligned}
$$