$$\begin{gathered} If\alpha ,\beta arerealand{\alpha }^2,-{\beta }^{2}aretherootsoftheequation \\ {a}^2{x}^2+x+(1-a^2)=0,a>1,then{\beta }^2= \end{gathered}$$

$\begin{array}{l}{\alpha }^2,-{\beta }^{2}arerootsof{a}^2{x}^2+x+(1-a^2)=0\\ \therefore {\alpha }^2-{\beta }^2=\frac{-1}{a^2},{\alpha }^2(-{\beta }^2)=\frac{1-a^2}{a^2}\\ ,-{\alpha }^2{\beta }^2=\frac{1-a^2}{a^2}\\ ({\alpha }^2+{\beta }^2)={({\alpha }^2-{\beta }^2)}^2+4{\alpha }^2{\beta }^2\\ =\frac{1}{a^4}+4\left(\frac{a^2-1}{a^2}\right)\\ ={(2-\frac{1}{a^2})}^2\\ {\alpha }^2+{\beta }^2=2-\frac{1}{a^2}\\ \therefore {\beta }^2=\frac{1}{2}[({\alpha }^2+{\beta }^2)-({\alpha }^2-{\beta }^2)]=\frac{1}{2}[2-\frac{1}{a^2}+\frac{1}{a^2}]=1\end{array}$