A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
Let eccentricity of\[{{x}^{2}}-{{y}^{2}}se{{c}^{2}}\alpha =5\]is\[{{e}_{1}}\]and eccentricity of \[{{x}^{2}}se{{c}^{2}}\alpha +{{y}^{2}}=25\]is\[{{e}_{2}},\]then \[{{e}_{1}}=\sqrt{3}{{e}_{2}}\] (given) \[\Rightarrow \] \[e_{1}^{2}=3e_{2}^{2}\] \[\Rightarrow \] \[\frac{5+5{{\cos }^{2}}\alpha }{5}=3.\frac{25-25{{\cos }^{2}}\alpha }{25}\] \[\Rightarrow \] \[1+{{\cos }^{2}}\alpha =3-3{{\cos }^{2}}\alpha \] \[\Rightarrow \] \[4{{\cos }^{2}}\alpha =2\] \[\Rightarrow \] \[{{\cos }^{2}}\alpha =\frac{1}{2}\] \[\Rightarrow \] \[\cos \alpha =\pm \frac{1}{\sqrt{2}}\Rightarrow \alpha =\frac{\pi }{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
