A) \[e_{1}^{2}+e_{2}^{2}=2\]
B) \[e_{1}^{2}+e_{2}^{2}=4\]
C) \[{{e}_{1}}+{{e}_{2}}=4\]
D) \[{{e}_{1}}+{{e}_{2}}=\sqrt{2}\]
Correct Answer: B
Solution :
Given equation can be rewritten as \[{{x}^{2}}-{{y}^{2}}=\frac{25}{3}\] Here, \[{{a}^{2}}=1,{{b}^{2}}=1\] \[\therefore \] s\[{{e}_{1}}=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+1}=\sqrt{2}\] The equation of conjugate hyperbola is \[-{{x}^{2}}+{{y}^{2}}=\frac{25}{3}\] \[\therefore \] \[{{e}_{2}}=\sqrt{1+\frac{{{a}^{2}}}{{{b}^{2}}}}=\sqrt{1+1}=\sqrt{2}\] \[\therefore \] \[e_{1}^{2}+e_{2}^{2}={{(\sqrt{2})}^{2}}+{{(\sqrt{2})}^{2}}=4\]
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