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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

The eccentricity of an ellipse , with its centre at origin is $\large\frac{1}{2}$, If one of the directrices is $x=4$ then the equation of the ellipse is

$\begin{array}{1 1}(A)\;3x^2+4y^2=1 \\(B)\;3x^2+4y^2=12 \\(C)\;4x^2+3y^2=12 \\(D)\;4x^2+3y^2=1 \end{array}$

1 Answer

Let the equation of ellipse be $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
Here $ a > b$ because the directrix parallel to y axis.
$b^2=a^2 (1-e^2)$ with $e =\large\frac{1}{2}$
=> $b^2= \large\frac{3}{4} $$a^2$
But $\large \frac{a}{e}$$=4$
=> $a=2$
Required ellipse is $ \large\frac{x^2}{4}+\frac{y^2}{3}$$=1$
=> $3x^2+4y^2=12$
Hence B is the correct answer.
answered Apr 10, 2014 by meena.p
 

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