In an ellipse,the distance from one of the fo | vedclass

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(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S_1(ae, 0)$ and $S_2(-ae, 0)$. The ends of the major axis are $A(a, 0)$

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$\frac{1}{8}$

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(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S_1(ae, 0)$ and $S_2(-ae, 0)$. The ends of the major axis are $A(a, 0)$ and $A'(-a, 0)$. The ends of the minor axis are $B(0, b)$ and $B'(0, -b)$.Given the distance from focus $S_1$ to the corresponding end of the major axis $A$ is $a - ae = 4 - \sqrt{7}$.Given the distance from focus $S_1$ to the end of the minor axis $B$ is $\sqrt{(ae)^2 + b^2} = 4$. Since $b^2 = a^2(1 - e^2)$,we have $\sqrt{(ae)^2 + a^2 - a^2e^2} = 4$,which implies $a = 4$.Substituting $a = 4$ into $a(1 - e) = 4 - \sqrt{7}$,we get $4(1 - e) = 4 - \sqrt{7}$,so $1 - e = 1 - \frac{\sqrt{7}}{4}$,which gives $e = \frac{\sqrt{7}}{4}$.Then $b^2 = a^2(1 - e^2) = 16(1 - \frac{7}{16}) = 16(\frac{9}{16}) = 9$,so $b = 3$.The distance $S_1S_2 = 2ae = 2(4)(\frac{\sqrt{7}}{4}) = 2\sqrt{7}$.In $	riangle BS_1S_2$,$BS_1 = BS_2 = 4$ and $S_1S_2 = 2\sqrt{7}$.Using the law of cosines,$\cos 	heta = \frac{BS_1^2 + BS_2^2 - S_1S_2^2}{2(BS_1)(BS_2)} = \frac{4^2 + 4^2 - (2\sqrt{7})^2}{2(4)(4)} = \frac{16 + 16 - 28}{32} = \frac{4}{32} = \frac{1}{8}$.

In an ellipse,the distance from one of the foci to its corresponding end of the major axis is $4-\sqrt{7}$ and the distance from the same focus to one end of the minor axis is $4$. Then the cosine of the angle subtended by the line segment joining its foci at one end of its minor axis is

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