A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{\sqrt{3}}\]
D) None of these
Correct Answer: C
Solution :
Let the ellipse be\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. Since,\[y=x\] and \[3y=-2x\] is a pair of conjugate diameters. Therefore, \[{{m}_{1}}{{m}_{2}}=-\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \] \[(1)\left( -\frac{2}{3} \right)=-\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \] \[2{{a}^{2}}=3{{b}^{2}}\] \[\Rightarrow \] \[2{{a}^{2}}=3{{a}^{2}}(1-{{e}^{2}})\] \[\Rightarrow \] \[2=3(1-{{e}^{2}})\] \[\Rightarrow \] \[{{e}^{2}}=\frac{1}{3}\] \[\Rightarrow \] \[e=\frac{1}{\sqrt{3}}\]
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