Annoying conics are annoying

The area of the ellipse is phi/2, so the unshaded area is phi/4. The shaded area is phi - phi/4 = 0.75 phi

Here you go @Anany Prakhar .

Let's first start by finding the coordinates of the vertices of the two parabolas that make up the whole hyperbola. Seeing how this has an axis of symmetry at $y=0$ , we plug in to the equation given, which produces $(\pm{1}, 0)$ . Thus, the diameter of the circle inscribed is $2$ . From here, we can deduce that the area of this circle is simply $\pi$ .

Now, let's examine the ellipse. Its major axis is the diameter of the larger circle that we just looked at. From the information we've been given, the minor axis is half of the major axis, which is thus $1$ . Now, we can plug into our formula for area $AB\pi$ where $A$ and $B$ are the maximum and minimum distance from the perimeter to the center of the ellipse. Thus, we have $0.5\pi$ as the area of the ellipse.

To solve for the area of the smallest circle, we see that its diameter is the minor axis of the ellipse described above. Thus, its area is $0.25\pi$ .

With our areas in hand, we can first find the area of non-ellipse section of the largest circle, which is $\pi-0.5\pi \Longrightarrow 0.5\pi$ . Then, we'll add the area of the smallest circle, $0.25\pi$ to get $0.75\pi$ . From here, our desired answer is simply $\boxed{0.75}$ .