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Question

The foci of an ellipse are $$( \pm 2,0 )$$ and its eccentricity is $$\dfrac { 1 } { 2 }$$ . Find its equation. 


Solution

Given eccentricity of ellipse $$ e=\dfrac{1}{2}$$

and focii = $$(\pm 2,0)$$

we know $$ e = \dfrac{c}{a}$$ where c is distance of focus from center.

from this $$c = 2.$$

hence a = $$\dfrac{2}{\dfrac 12} = 4 $$

now eccentricity $$ e = \sqrt{1-\dfrac{b^{2}}{a^{2}}}=\dfrac{1}{2}$$

$$ \dfrac{b^{2}}{a^{2}} = 1-\dfrac{1}{4} = \dfrac{3}{4}$$

$$ b^{2} = \dfrac{3}{4}\times (4)^{2} = 12 $$

Hence equation of ellipse is

$$ \Rightarrow \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} = 1$$

$$ \Rightarrow \dfrac{x^{2}}{16}+\dfrac{y^{2}}{12} = 1 $$ Ans.

Maths

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