A) \[e>\sqrt{3}\]
B) \[1<e<\frac{2}{\sqrt{3}}\]
C) \[e=\frac{2}{\sqrt{3}}\]
D) \[e>\frac{2}{\sqrt{3}}\]
Correct Answer: D
Solution :
[d] Let the length of the double ordinate be \[2\ell .\] |
\[\therefore \,\,\,\,AB=2\ell \] and \[AM=BM=\ell \] |
Clearly ordinate of point A is \[\ell \]. |
The abscissa of the point A is given by |
\[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{\ell }^{2}}}{{{b}^{2}}}=1\Rightarrow x=\frac{a\sqrt{{{b}^{2}}+{{\ell }^{2}}}}{b}\] |
\[\therefore \] A is \[\left( \frac{a\sqrt{{{b}^{2}}+{{\ell }^{2}}}}{b},\ell \right)\] |
Since \[\Delta OAB\] is equilateral triangle, therefore |
\[OA=AB=OB=2\ell .\] |
Also, \[O{{M}^{2}}+A{{M}^{2}}=O{{A}^{2}}\] |
\[\therefore \frac{{{a}^{2}}({{b}^{2}}+{{\ell }^{2}})}{{{b}^{2}}}+{{\ell }^{2}}=4{{\ell }^{2}}\] |
We get \[{{\ell }^{2}}=\frac{{{a}^{2}}{{b}^{2}}}{3{{b}^{2}}-{{a}^{2}}}\] |
Since \[{{\ell }^{2}}>0\therefore \frac{{{a}^{2}}{{b}^{2}}}{3{{b}^{2}}-{{a}^{2}}}>0\Rightarrow 3{{b}^{2}}-{{a}^{2}}>0\] |
\[\Rightarrow 3{{a}^{2}}({{e}^{2}}-1)-{{a}^{2}}>0\Rightarrow e>\frac{2}{\sqrt{3}}\] |
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