A) \[-1\]
B) 0
C) 1
D) 2
Correct Answer: C
Solution :
Let the hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Eccentricity, \[e=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}\] \[\Rightarrow \] \[\frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{({{a}^{2}}+{{b}^{2}})}\] Conjugate hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1\] Eccentricity, \[{{e}_{1}}=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}}\] \[\Rightarrow \] \[\frac{1}{e_{1}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] Now, \[\frac{1}{{{e}^{2}}}+\frac{1}{e_{1}^{2}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
