JEE MAIN - Mathematics (2004 - No. 18)
The eccentricity of an ellipse, with its centre at the origin, is $${1 \over 2}$$. If one of the directrices is $$x=4$$, then the equation of the ellipse is :
$$4{x^2} + 3{y^2} = 1$$
$$3{x^2} + 4{y^2} = 12$$
$$4{x^2} + 3{y^2} = 12$$
$$3{x^2} + 4{y^2} = 1$$
Explanation
$$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$
$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$
$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$
Equation of elhipe is
$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$
$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$
$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$
Equation of elhipe is
$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$
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