A) \[{{\left( \frac{1}{e} \right)}^{2}}+{{\left( \frac{1}{e'} \right)}^{2}}=1\]
B) \[\frac{1}{e}+\frac{1}{e'}=1\]
C) \[{{\left( \frac{1}{e} \right)}^{2}}+{{\left( \frac{1}{e'} \right)}^{2}}=0\]
D) \[\frac{1}{e}+\frac{1}{e'}=2\]
Correct Answer: A
Solution :
Let hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] .....(i) Then its conjugate will be, \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1\] .....(ii) If \[e\] is eccentricity of hyperbola (i), then \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)\] or \[\frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{({{a}^{2}}+{{b}^{2}})}\] .....(iii) Similarly if e' is eccentricity of conjugate (ii), then \[{{a}^{2}}={{b}^{2}}(e{{'}^{2}}-1)\] or \[\frac{1}{e{{'}^{2}}}=\frac{{{b}^{2}}}{({{a}^{2}}+{{b}^{2}})}\] .....(iv) Adding (iii) and (iv), \[\frac{1}{{{(e')}^{2}}}+\frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1.\]
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