question_answer

                If a and c are positive real numbers and the ellipse\[\frac{{{x}^{2}}}{4{{c}^{2}}}+\frac{{{y}^{3}}}{{{c}^{2}}}=1\]has four distinct points in common with the circle \[{{x}^{2}}+{{y}^{2}}=9{{a}^{2}},\] then                   JEE Main Online Paper (Held On 09 April 2013)

A)                 \[9a-9{{a}^{2}}-2{{c}^{2}}<0\]                    

B)                 \[6ac+9{{a}^{2}}-2{{c}^{2}}<0\]                

C)                 \[9ac-9{{a}^{2}}-2{{c}^{2}}>0\]                

D)                 \[6ac+9{{a}^{2}}-2{{c}^{2}}>0\]                

Correct Answer: C

Solution :

                                Radius of the circle having SS' as diameter is \[r=ae\] if it cuts an ellipse then r > b                 \[ae>b\Rightarrow \,{{e}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\]                 \[{{e}^{2}}>1-{{e}^{2}}\]                 \[{{e}^{2}}>\frac{1}{2}\]                 \[e>\frac{1}{\sqrt{2}}\Rightarrow \,e\in \,\,\left( \frac{1}{\sqrt{2}},\,1 \right)\]