A) \[\pi /6\]
B) \[\pi /4\]
C) \[\pi /3\]
D) \[\pi /2\]
Correct Answer: B
Solution :
[b] For the hyperbola \[\frac{{{x}^{2}}}{5}-\frac{{{y}^{2}}}{5{{\cos }^{2}}\alpha }=1\] We have \[{{e}_{1}}^{2}=1+\frac{{{b}^{2}}}{{{a}^{2}}}=1+\frac{5{{\cos }^{2}}\alpha }{5}=1+{{\cos }^{2}}\alpha \] For the ellipse \[\frac{{{x}^{2}}}{25{{\cos }^{2}}\alpha }+\frac{{{y}^{2}}}{25}=1\] We have \[{{e}_{2}}^{2}=1-\frac{25{{\cos }^{2}}\alpha }{25}={{\sin }^{2}}\alpha \] Given that \[{{e}_{1}}=\sqrt{3}{{e}_{2}}\] \[\therefore {{e}_{1}}^{2}=3{{e}_{2}}^{2}\] Or \[1+{{\cos }^{2}}\alpha =3{{\sin }^{2}}\alpha \] Or \[2=4{{\sin }^{2}}\alpha \] Or \[\sin \alpha =\frac{1}{\sqrt{2}}\]
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