A) 1
B) 2
C) 3
D) None of these
Correct Answer: A
Solution :
\[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] Þ \[{{e}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}\] \[{{e}_{1}}=\sqrt{1+\frac{{{a}^{2}}}{{{b}^{2}}}}\] Þ \[e_{1}^{2}=\frac{{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}}\] Þ \[\frac{1}{e_{1}^{2}}+\frac{1}{{{e}^{2}}}=1\].You need to login to perform this action.
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