A) \[\frac{5}{4}\]
B) \[\frac{5}{3}\]
C) \[\frac{4}{3}\]
D) \[\frac{9}{4}\]
Correct Answer: B
Solution :
[b]: Clearly, for the first curve \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)={{a}^{2}}\left( \frac{25}{16}-1 \right)=\frac{9{{a}^{2}}}{16}\] If\[{{e}_{1}}\] is the eccentricity of the second curve, then \[{{a}^{2}}={{b}^{2}}(e_{1}^{2}-1)\] or \[\frac{16}{9}=e_{1}^{2}-1\] \[\Rightarrow \]\[e_{1}^{2}=\frac{16}{9}+1=\frac{25}{9}\]\[\Rightarrow \]\[{{e}_{1}}=\frac{5}{3}\].You need to login to perform this action.
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