and\[e\]be the eccentricities of a hyperbola and its conjugate, then\[\frac{1}{{{e}^{2}}}+\frac{1}{e{{}^{2}}}\]is equal to
A) 0
B) 1
C) 2
D) None of these
Correct Answer: B
Solution :
Let a hyperbola, \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] ?(i) and its conjugate hyperbola \[-\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Eccentricities of the hyperbolas (i) and (ii) are given by \[{{e}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}\] and\[e{{}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}\]respectively Now, \[\frac{1}{{{e}^{2}}}+\frac{1}{e{{}^{2}}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\]
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