7. Let the three sides of $\triangle A B C$ be $a, b, c$, and $b>\max \{a, c\}$. There are three non-zero real numbers $x_{0}, y_{0}, z_{0}$ such that the line $a x+b y+c=0$ passes through the point $\left(\frac{z_{0}}{x_{0}}, \frac{2 y_{0}}{x_{0}}\right)$, and the point $\left(\frac{z_{0}}{y_{0}}, \frac{x_{0}}{y_{0}}\right)$ lies on the ellipse $x^{2}+\frac{y^{2}}{4}=1$. Then the maximum value of $\tan B \cdot \cot C$ is . $\qquad$
7. $\frac{5}{3}$.

From the problem, we know
$$
4 z_{0}^{2}+x_{0}^{2}=4 y_{0}^{2}, a z_{0}+2 b y_{0}+c x_{0}=0 \text {. }
$$

Then $4 b^{2} y_{0}^{2}=\left(a z_{0}+c x_{0}\right)^{2}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{c}
\left(x_{0}^{2}+4 z_{0}^{2}\right)\left(c^{2}+\frac{a^{2}}{4}\right) \geqslant\left(c x_{0}+a z_{0}\right)^{2}, \\
4 y_{0}^{2}\left(c^{2}+\frac{a^{2}}{4}\right) \geqslant 4 b^{2} y_{0}^{2} \Rightarrow c^{2}+\frac{a^{2}}{4} \geqslant b^{2} .
\end{array}
$$

Since $a^{2}+c^{2}>c^{2}+\frac{a^{2}}{4} \geqslant b^{2}(b>a, b>c)$,
thus, $\triangle A B C$ is an acute triangle.
By the Law of Sines, we have
$$
\begin{array}{l} 
\sin ^{2} C+\frac{1}{4} \sin ^{2} A \geqslant \sin ^{2} B \\
\Rightarrow \frac{1}{4} \sin ^{2} A \geqslant \sin ^{2} B-\sin ^{2} C \\
=\sin (B+C) \cdot \sin (B-C) \\
=\sin A \cdot \sin (B-C) \\
\Rightarrow \sin A \geqslant 4 \sin (B-C) \\
\Rightarrow \sin B \cdot \cos C+\sin C \cdot \cos B \\
\quad \geqslant 4 \sin B \cdot \cos C-4 \sin C \cdot \cos B \\
\Rightarrow 5 \sin C \cdot \cos B \geqslant 3 \sin B \cdot \cos C \\
\Rightarrow \tan B \cdot \cot C \leqslant \frac{5}{3} .
\end{array}
$$

When $\frac{x_{0}}{c}=\frac{4 z_{0}}{a}$, the equality holds.