Transformation Patterns

First, consider a unit circle with two tangent parallel lines. Then, pick a point $A$ $x$ away from $B$ on the line on top and draw another tangent line that passes trough it. This line intersects the bottom line $y$ meters away from $F$ . It is preety easy to check with some angle hunting that $ABCD$ is similar to $CDEF$ , and so, $\frac{y}{1}=\frac{1}{x}$ . If you place the center of that circle at point $(0,1)$ you have almost the setup in the problem, the only difference is that instead of making point $(x,1)$ to $(\frac{1}{x},0)$ , it maps point $(x,2)$ to $(\frac{1}{x}),2$ , so all we need to do is "compress" everything vertically by a factor of $2$ . Since the circle has area $\pi$ , the region we want has area $\frac{\pi}{2}$

For a circle of radius $1$ centered at $0,1$ , a tangent passing through $(x,2)$ will pass through $(\frac{1}{x},0)$ . Hence, for an ellipse of major & minor axes of $1, \frac{1}{2}$ , a tangent passing through $(x,1)$ will pass through $(\frac{1}{x},0)$ , and so the area of this ellipse is $\frac{\pi}{2}$ .

Okay, I just now see Cardoso's solution, same idea.