JEE Advance - Mathematics (1996 - No. 25)
An ellipse has eccentricity $${1 \over 2}$$ and one focus at the point $$P\left( {{1 \over 2},1} \right)$$. Its one directrix is the common tangent, nearer to the point $$P$$, to the circle $${x^2} + {y^2} = 1$$ and the hyperbol;a $${x^2} - {y^2} = 1$$. The equation of the ellipse, in the standard form, is ............
${{{{\left( {x - {1 \over 3}} \right)}^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} + {{{{\left( {y - 1} \right)}^2}} \over {{{\left( {{1 \over {2\sqrt 3 }}} \right)}^2}}} = 1$
${{{{\left( {x + {1 \over 3}} \right)}^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} + {{{{\left( {y - 1} \right)}^2}} \over {{{\left( {{1 \over {2\sqrt 3 }}} \right)}^2}}} = 1$
${{{{\left( {x - {1 \over 3}} \right)}^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}} + {{{{\left( {y - 1} \right)}^2}} \over {{{\left( {{1 \over {2\sqrt 3 }}} \right)}^2}}} = 1$
${{{{\left( {x - {1 \over 3}} \right)}^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} + {{{{\left( {y + 1} \right)}^2}} \over {{{\left( {{1 \over {2\sqrt 3 }}} \right)}^2}}} = 1$
${{{{\left( {x - {1 }} \right)}^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} + {{{{\left( {y - 1} \right)}^2}} \over {{{\left( {{1 \over {2\sqrt 3 }}} \right)}^2}}} = 1$
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