\(I _{2}=4 I _{1}\)
\(I _{\max }= I _{1}+4 I _{1}+2 \sqrt{ I _{1} 4 I _{1}}=9 I _{1}\)
\(I_{\min }=I_{1}+4 I_{1}-2 \sqrt{I_{1} 4 I_{1}}=I_{1}\)
\(\therefore \frac{9 I _{1}+ I _{1}}{9 I _{1}- I _{1}}=\frac{10}{8}=\frac{5}{4}=\frac{2 \alpha+1}{\beta+1}\)
\(\alpha=2 \beta=1\)
\(\therefore \frac{\alpha}{\beta}=\frac{2}{1}=2\)