A) 10: 8
B) 9:1
C) 4:1
D) 2:1
Correct Answer: C
Solution :
\[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}=\frac{9}{1}\] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\frac{9}{1}}\] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{1}\] \[\frac{{{l}_{\max }}}{{{l}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}=\frac{{{(3+1)}^{2}}}{{{(3-1)}^{2}}}\] \[=\frac{{{(4)}^{2}}}{{{(2)}^{2}}}=\frac{16}{4}=\frac{4}{1}\] \[{{l}_{\max }}:{{l}_{\min }}=4:1\]
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