A) 0
B) 1
C) 2
D) none of these
Correct Answer: B
Solution :
Key Idea: If e is eccentricity f hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1,\]then \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}.\] Since, e is eccentricity of hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\therefore \] \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] \[\Rightarrow \] \[{{e}^{2}}=1+\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}\] and e is eccentricity of hyperbola \[\frac{{{x}^{2}}}{{{b}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}}=1\] \[\therefore \] \[e=\sqrt{1+\frac{{{a}^{2}}}{{{b}^{2}}}}\] \[\Rightarrow \] \[{{(e)}^{2}}=1+\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}\] \[\therefore \] \[\frac{1}{{{e}^{2}}}+\frac{1}{{{(e)}^{2}}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\]
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