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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 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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$\frac{1}{8}$
$\begin{aligned} & \text { } S_1 A=4-\sqrt{7} \Rightarrow|a e-a|=4-\sqrt{7} \\ & a^2(1-e)^2=(4-\sqrt{7})^2 \\ & B S_1=4\end{aligned}$
$\begin{aligned} & (a e)^2+b^2=16 \Rightarrow(a e)^2+a^2-(a e)^2=16 \\ & \Rightarrow \quad a^2=16 \Rightarrow a=4 \\ & \therefore \quad 4(1-e)=4-\sqrt{7} \\ & \Rightarrow \quad 1-\left(\frac{4-\sqrt{7}}{4}\right)=e \Rightarrow e=\frac{\sqrt{7}}{4} \\ & \therefore \quad b=a \sqrt{1-e^2}=4 \times \frac{3}{4}=3 \\ & \quad B S_1=B S_2=4 \\ & S_1 S_2=2 a e=2 \sqrt{7} \\ & \cos \theta=\frac{B S_1^2+B S_2^2-\left(S_1 S_2\right)^2}{2\left(B S_1\right)\left(B S_2\right)}=\frac{16+16-28}{2(4)(4)} \\ & \therefore \quad \cos \theta=\frac{1}{8} .\end{aligned}$
$\begin{aligned} & (a e)^2+b^2=16 \Rightarrow(a e)^2+a^2-(a e)^2=16 \\ & \Rightarrow \quad a^2=16 \Rightarrow a=4 \\ & \therefore \quad 4(1-e)=4-\sqrt{7} \\ & \Rightarrow \quad 1-\left(\frac{4-\sqrt{7}}{4}\right)=e \Rightarrow e=\frac{\sqrt{7}}{4} \\ & \therefore \quad b=a \sqrt{1-e^2}=4 \times \frac{3}{4}=3 \\ & \quad B S_1=B S_2=4 \\ & S_1 S_2=2 a e=2 \sqrt{7} \\ & \cos \theta=\frac{B S_1^2+B S_2^2-\left(S_1 S_2\right)^2}{2\left(B S_1\right)\left(B S_2\right)}=\frac{16+16-28}{2(4)(4)} \\ & \therefore \quad \cos \theta=\frac{1}{8} .\end{aligned}$
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