$2 \cdot 70$ For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
(i) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
(ii) There exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, such that
$$\begin{array}{l}
{\left[(1-x) \sin \theta-\sqrt{a-x^{2} \cos ^{2} \theta}\right]^{2}+\left[x \cos \theta-\sqrt{a-(1-x)^{2} \sin ^{2} \theta}\right]^{2}} \\
\leqslant a
\end{array}$$
[Solution] From (i) we get
$$\sqrt{a}>\frac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}$$

Without loss of generality, assume $\quad \frac{a}{\sin ^{2} \theta}+\frac{a}{\cos ^{2} \theta} \leqslant 1$.
(ii) is equivalent to: there exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, satisfying
$$2(1-x) \sin \theta \sqrt{a-x^{2} \cos ^{2} \theta}+2 x \cos \theta \sqrt{a-(1-x)^{2} \sin ^{2} \theta} \geqslant a,$$

which is equivalent to
$$2 \sin \theta \cos \theta\left[(1-x) \sqrt{\frac{a}{\cos ^{2} \theta}-x^{2}}+x \sqrt{\frac{a}{\sin ^{2} \theta}-(1-x)^{2}}\right] \geqslant a .$$

First, we prove a lemma: Let $o1, p^{2}+q^{2} \leqslant 1$, $f(x)=(1-x) \sqrt{p^{2}-x^{2}}+x \sqrt{q^{2}-(1-x)^{2}}(1-q \leqslant x \leqslant p)$. Then when $\sqrt{p^{2}-x^{2}}=\sqrt{q^{2}-(1-x)^{2}}$, i.e., $x=\frac{p^{2}-q^{2}+1}{2} \in[1-q, p]$, $f(x)$ reaches its maximum value.

Proof of the lemma:
Since $1-q \leqslant x \leqslant p$, we can let
$$x=p \sin \alpha, 1-x=q \sin \beta, 00$$
(4) has roots
$$\begin{aligned}
& \frac{\sin ^{4} \theta \cos ^{4} \theta}{1-3 \sin ^{2} \theta \cos ^{2} \theta}\left[\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta} \pm\right. \\
& \left.\sqrt{\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\right)^{2}-\frac{1}{\cos ^{4} \theta}-\frac{1}{\sin ^{4} \theta}+\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}}\right] \\
= & \frac{\sin ^{2} \theta \cos ^{2} \theta}{1 \mp \sqrt{3} \sin \theta \cos \theta}
\end{aligned}$$

Therefore, from (4) we get
$$\begin{array}{l}
\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} \leqslant a \leqslant \frac{\sin ^{2} \theta \cos ^{2} \theta}{1-\sqrt{3} \sin \theta \cos \theta} \\
\frac{\sin ^{2} \theta \cos ^{2} \theta}{(\sin \theta+\cos \theta)^{2}}<\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta}
\end{array}$$

Thus, when
$a=\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta}$, (1) is satisfied, hence the required
$a=\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta}$.