11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • 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 ellipse is [AIEEE 2004]

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

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

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

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

    Correct Answer: B

    Solution :

               Since directrix is parallel to y-axis, hence axes of the ellipse are parallel to x-axis.            Let the equation of the ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\],  \[(a>b)\]            \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=1-{{e}^{2}}=1-\frac{1}{4}\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{3}{4}\].            Also, one of the directrices is \[x=4\]            \[\Rightarrow \]\[\frac{a}{e}=4\Rightarrow a=4e=4.\frac{1}{2}=2\]; \[{{b}^{2}}=\frac{3}{4}{{a}^{2}}=\frac{3}{4}.4=3\]            \[\therefore \] Required ellipse is \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\] or \[3{{x}^{2}}+4{{y}^{2}}=12\].


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