Task B-4.6. Determine the equations of all tangents to the ellipse $x^{2}+4 y^{2}=20$ such that the points of tangency bisect the segments that the coordinate axes cut off on these tangents. Calculate the area of the quadrilateral determined by these tangents.
## Second Solution. Sketch

The equation of the tangent is of the form $y=k x+l$. The intersections of the tangent with the coordinate axes are $M\left(-\frac{l}{k}, 0\right), N(0, l)$. The coordinates of the midpoint are $P\left(-\frac{l}{2 k}, \frac{l}{2}\right)$.

(2 points)

The coordinates of the point of tangency on the ellipse are $\left(-\frac{a^{2} k}{l}, \frac{b^{2}}{l}\right)$.

If we equate the coordinates of the midpoint and the coordinates of the point of tangency, we get

$$
\begin{aligned}
\frac{l}{2}=\frac{b^{2}}{l} \quad \Rightarrow \quad l^{2}=2 b^{2}=2 \cdot 5=10 \quad \Rightarrow \quad l= \pm \sqrt{10} \\
-\frac{l}{2 k}=-\frac{a^{2} k}{l} \quad \Rightarrow \quad 2 k^{2} a^{2}=l^{2} \quad \Rightarrow \quad k^{2}=\frac{l^{2}}{2 a^{2}}=\frac{10}{40}=\frac{1}{4} \quad \Rightarrow \quad k= \pm \frac{1}{2}
\end{aligned}
$$

The equations of the tangents are

$$
\begin{aligned}
& y=\frac{1}{2} x+\sqrt{10} \\
& y=-\frac{1}{2} x+\sqrt{10} \\
& y=\frac{1}{2} x-\sqrt{10} \\
& y=-\frac{1}{2} x-\sqrt{10}
\end{aligned}
$$

The area of the resulting rhombus is $P=4 \cdot \frac{1}{2} \cdot 2 \sqrt{10} \cdot \sqrt{10}=40$.