A) 4
B) 3
C) 2
D) 1
Correct Answer: C
Solution :
Since, \[\frac{e}{2}\] and \[\frac{e'}{2}\] are the eccentricities of a hyperbola and its conjugate hyperbola. Then, \[\frac{1}{{{\left( \frac{e}{2} \right)}^{2}}}+\frac{1}{{{\left( \frac{e'}{2} \right)}^{2}}}=1\] \[\Rightarrow \] \[1=\frac{4}{{{e}^{2}}}+\frac{4}{e{{'}^{2}}}\] \[\Rightarrow \] \[4=\frac{{{e}^{2}}e{{'}^{2}}}{{{e}^{2}}+{{e}^{'2}}}\] ?(i)You need to login to perform this action.
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