9. 26 Given a real number $a$, find the largest real number $\lambda$ such that any rectangle in the plane with sides parallel to the $x$-axis and $y$-axis, if it contains the figure $G$ defined by the following inequalities:
$$\left\{\begin{array}{l}
y \leqslant -x^{2}, \\
y \geqslant x^{2}-2 x+a
\end{array}\right.$$

then its area is not less than $\lambda$.
[Solution] Clearly, the $x$-coordinates of the points in $G$ satisfy
$$2 x^{2}-2 x+a \leqslant 0 \text {. }$$

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 this, 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,
$$\lambda=(1-a) \sqrt{1-2 a}$$

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