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(A) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.Here,$a^2 = 4$ and $b^2 = 1$,so $a = 2

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$2 \cos \frac{\alpha-\beta}{2}$

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(A) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.Here,$a^2 = 4$ and $b^2 = 1$,so $a = 2$ and $b = 1$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.The coordinates of the extremities of a focal chord with eccentric angles $\alpha$ and $\beta$ are $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$.The condition for a chord joining $\alpha$ and $\beta$ to be a focal chord is $\cos \frac{\alpha-\beta}{2} = e \cos \frac{\alpha+\beta}{2}$.Substituting $e = \frac{\sqrt{3}}{2}$,we get $\cos \frac{\alpha-\beta}{2} = \frac{\sqrt{3}}{2} \cos \frac{\alpha+\beta}{2}$.Multiplying both sides by $2$,we get $2 \cos \frac{\alpha-\beta}{2} = \sqrt{3} \cos \frac{\alpha+\beta}{2}$.Thus,$\sqrt{3} \cos \frac{\alpha+\beta}{2} = 2 \cos \frac{\alpha-\beta}{2}$.

If $\alpha, \beta$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^2+4y^2=4$,then $\sqrt{3} \cos \frac{\alpha+\beta}{2} =$

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