Let F and F' be the foci of the ellipse frac{ | vedclass

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(A) The given equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ with $b < 2$. Here,$a^2 = 4$,so $a = 2$. The foci are $F(c, 0)$ and $F'(-c,

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$\frac{\sqrt{3}}{2}$ or $\frac{1}{2}$

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(A) The given equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ with $b Here,$a^2 = 4$,so $a = 2$. The foci are $F(c, 0)$ and $F'(-c, 0)$,and the end of the minor axis is $B(0, b)$. The area of $	riangle FBF'$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2c) 	imes b = bc$. Given $bc = \sqrt{3}$,so $b^2 c^2 = 3$,which implies $c^2 = \frac{3}{b^2}$. For an ellipse,$b^2 = a^2(1 - e^2) = a^2 - a^2 e^2 = a^2 - c^2$. Thus,$b^2 = 4 - c^2$,or $c^2 = 4 - b^2$. Substituting $c^2 = \frac{3}{b^2}$ into $c^2 = 4 - b^2$: $\frac{3}{b^2} = 4 - b^2$ $\Rightarrow 3 = 4b^2 - b^4$ $\Rightarrow b^4 - 4b^2 + 3 = 0$. Let $t = b^2$,then $t^2 - 4t + 3 = 0 \Rightarrow (t - 1)(t - 3) = 0$. So,$b^2 = 1$ or $b^2 = 3$. Case $1$: $b^2 = 1$. Then $c^2 = 4 - 1 = 3$,so $c = \sqrt{3}$. Eccentricity $e = \frac{c}{a} = \frac{\sqrt{3}}{2}$. Case $2$: $b^2 = 3$. Then $c^2 = 4 - 3 = 1$,so $c = 1$. Eccentricity $e = \frac{c}{a} = \frac{1}{2}$. Thus,the eccentricity is $\frac{\sqrt{3}}{2}$ or $\frac{1}{2}$.

Let $F$ and $F'$ be the foci of the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ $(b < 2)$ and $B$ is one end of the minor axis. If the area of the triangle $FBF'$ is $\sqrt{3}$ sq. units,then the eccentricity of the ellipse is

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