5. Given the family of lines $L: \frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1(a>b>0)$, then
A. If $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}<1$, then the maximum area of the triangle formed by the lines in $L$ is $\frac{3 \sqrt{3}}{4} a b$
C. If $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$, then the maximum area of the triangle formed by the lines in $L$ is $3 \sqrt{3} a b$
D. If $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$, then the lines in $L$ can form a square
From the concept of pole and polar, if $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}<1$, then the area of the triangle formed by the lines in $L$ is maximized when the triangle is inscribed in the ellipse, which can be converted to the inscribed triangle in a circle, and the maximum area at this time is $\frac{3 \sqrt{3}}{4} a b$, so option $B$ is correct; if $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$, then the triangle formed by the lines in $L$ is circumscribed around the ellipse, which can be converted to the circumscribed triangle in a circle, and the minimum area at this time is $3 \sqrt{3} a b$, so option $C$ is incorrect; as shown in the figure, if $E F G H$ is a square, let $A\left(x_{0}, y_{0}\right)$, then $k_{E F}=-\frac{b^{2} x_{0}}{a^{2} y_{0}}=-1 \Rightarrow x_{0}=k a^{2}, y_{0}=k b^{2}$, substituting into $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1 \Rightarrow k=\frac{1}{\sqrt{a^{2}+b^{2}}}$. It is known that when $A\left(\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}, \frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right)$, it satisfies the condition, so option $D$ is correct. Therefore, the correct answers are $B D$.