3. In the common area of the inner regions of the parabolas

$$
x^{2}=p^{2}+2 p y \text { and } x^{2}=p^{2}-2 p y
$$

inscribe an ellipse with the maximum area.
Solution. If the ellipse is located inside the region bounded by the parabolas

$$
x^{2}=p^{2}+2 p y \text { and } x^{2}=p^{2}-2 p y
$$

where, for example, $p>0$, as shown in the right diagram, then by translating the ellipse so that the intersection of the axes coincides with the coordinate origin, and then rotating around the coordinate origin so that the major axis of the ellipse falls on the $x$-axis, we will obtain an ellipse that is inside the mentioned region (Prove it!).

The equation of the desired ellipse with maximum area has the form

$$
\frac{x^{2}}{\lambda^{2} b^{2}}+\frac{y^{2}}{b^{2}}=1,
$$

where $0 < b$ and $\frac{p}{\lambda_{2}^{2}} < P_{1}$. The desired ellipse is

$$
\frac{x^{2}}{\frac{2 p^{2}}{3}}+\frac{y^{2}}{\frac{2 p^{2}}{9}}=1
$$