JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        The eccentricity of an ellipse with its centre at the origin, is\[\frac{1}{2}\]. If one of the directrices is \[x=4,\]then the equation of the elapse is:

    A)  \[3{{x}^{2}}+4{{y}^{2}}=1\]        

    B)  \[3{{x}^{2}}+4{{y}^{2}}=12\]

    C)  \[4{{x}^{2}}+3{{y}^{2}}=12\]     

    D)  \[4{{x}^{2}}+3{{y}^{2}}=1\]

    Correct Answer: B

    Solution :

                    Since directrix is\[x=4,\]then major axis of an ellipse is along\[x-\]axis. \[\therefore \]  \[\frac{a}{e}=4\Rightarrow a=4e\Rightarrow 4\times \frac{1}{2}\] \[\Rightarrow \]                               \[a=2\]                 \[\left( \because e=\frac{1}{2} \right)\] Also we know that\[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\]                 \[{{b}^{2}}=4\left( 1-\frac{1}{4} \right)=4\times \frac{3}{4}\]                 \[{{b}^{2}}=3\] \[\therefore \]Equation of ellipse is \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\] \[\Rightarrow \]                               \[3{{x}^{2}}+4{{y}^{2}}=12\]


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