Let frac{x^2}{a^2} + frac{y^2}{b^2} = 1 (b

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(C) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b < a$. The coordinates are $C = (0, b)$,$F_1 = (-ae, 0)$,and $B = (a, 0)$.Since $

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$\frac{\sqrt{5}-1}{2}$

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(C) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b Since $\angle F_1CB = 90^{\circ}$,the product of the slopes of $CF_1$ and $CB$ is $-1$.Slope of $CF_1 = \frac{0 - b}{-ae - 0} = \frac{-b}{-ae} = \frac{b}{ae}$.Slope of $CB = \frac{0 - b}{a - 0} = \frac{-b}{a}$.Therefore,$(\frac{b}{ae}) 	imes (\frac{-b}{a}) = -1$.$\frac{b^2}{a^2e} = 1 \Rightarrow b^2 = a^2e$.Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2e = a^2(1 - e^2)$.$e = 1 - e^2 \Rightarrow e^2 + e - 1 = 0$.Solving for $e$ using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since eccentricity $e > 0$,we have $e = \frac{\sqrt{5} - 1}{2}$.

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(b < a)$ be an ellipse with major axis $AB$ and minor axis $CD$. Let $F_1$ and $F_2$ be its two foci,with $A, F_1, F_2, B$ in that order on the segment $AB$. Suppose $\angle F_1CB = 90^{\circ}$. The eccentricity of the ellipse is:

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