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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