$9 \cdot 26$ Given a real number $a$, find the largest real number $\lambda$, such that the area of any rectangle in the plane with sides parallel to the $x$-axis and $y$-axis, which contains the figure $G$ defined by the following inequalities:
$$
\left\{\begin{array}{l}
y \leqslant -x^{2}, \\
y \geqslant x^{2}-2 x+a,
\end{array}\right.
$$
is not less than $\lambda$.
[Solution] Clearly, the $x$-coordinates of the points in $G$ satisfy
$$
2 x^{2}-2 x+a \leqslant 0 .
$$

Thus, when $a>\frac{1}{2}$, $G$ is an empty set. When $a=\frac{1}{2}$, $x=\frac{1}{2}$, from which we immediately get that $G$ is a single point $\left(\frac{1}{2},-\frac{1}{4}\right)$. Therefore, when $a \geqslant \frac{1}{2}$, $\lambda=0$.
If $a<\frac{1}{2}$, from (1) we know that the $x$-coordinates of the points in $G$ satisfy
$$
\frac{1}{2}-\frac{1}{2} \sqrt{1-2 a} \leqslant x \leqslant \frac{1}{2}+\frac{1}{2} \sqrt{1-2 a} .
$$

There are two cases:
(1) $0<a<\frac{1}{2}$. Since $0<\frac{1}{2}-\frac{1}{2} \sqrt{1-2 a}<\frac{1}{2}+\frac{1}{2} \sqrt{1-2 a}<$ 1, the $y$-coordinates of the points in $G$ satisfy
$$
-\left(\frac{1}{2}+\frac{1}{2} \sqrt{1-2 a}\right)^{2} \leqslant y \leqslant-\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-2 a}\right)^{2},
$$

from which we get $\lambda=1-2 a$.
(2) $a<0$. Since $\frac{1}{2}-\frac{1}{2} \sqrt{1-2 a} \leqslant 0,1 \leqslant \frac{1}{2}+\frac{1}{2} \sqrt{1-2 a}$, the $y$-coordinates of the points in $G$ satisfy
$$
a-1 \leqslant y \leqslant 0 .
$$

Thus,
$$
\lambda=(1-a) \sqrt{1-2 a} \text {. }
$$

In summary,
$$
\lambda=\left\{\begin{array}{ll}
0, & a \geqslant \frac{1}{2} . \\
1-2 a, & 0<a<\frac{1}{2} . \\
(1-a) \sqrt{1-2 a}, & a<0 .
\end{array}\right.
$$