The eccentricity of an ellipse, with its cent | vedclass

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Question

(b) 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{{

Vedclass · Accepted Answer

$3{x^2} + 4{y^2} = 12$

Vedclass · Answer

(b) 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$

$\Rightarrow a = 4e = 4.\frac{1}{2} = 2$;

${b^2} = \frac{3}{4}{a^2} = \frac{3}{4}.4 = 3$

$	herefore $ Required ellipse is $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1$

or $3{x^2} + 4{y^2} = 12$.

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

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