A) \[\frac{4}{3}\]
B) 2
C) \[\sqrt{3}\]
D) \[\frac{2}{\sqrt{3}}\]
Correct Answer: D
Solution :
\[e_{1}^{2}\,=1+\frac{{{b}^{2}}}{{{a}^{2}}}\,=1+\frac{12}{4}\,=4\]\[\Rightarrow \,\,{{e}_{1}}=2;\] now \[\frac{1}{e_{1}^{2}}\,+\frac{1}{e_{2}^{2}}=1\] \[\frac{1}{e_{1}^{2}}\,=1-\frac{1}{4}=\frac{3}{4}\Rightarrow \,e_{2}^{2}\,=\frac{4}{3}\,\Rightarrow \,\,{{e}_{2}}=\,\frac{2}{\sqrt{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
