S^{prime} is the focus of the ellipse frac{x^ | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{b^2} = 1$,where $a^2 = 25$,so $a = 5$.Given the distance between the foci is $2ae = 8$

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$\frac{\pi}{3}$

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(B) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{b^2} = 1$,where $a^2 = 25$,so $a = 5$.Given the distance between the foci is $2ae = 8$,we have $2(5)e = 8$,which gives $e = \frac{4}{5}$.The coordinates of the focus $S^{\prime}$ on the negative $X$-axis are $(-ae, 0) = (-4, 0)$.$A$ point $P$ on the ellipse is given by $(a \cos 	heta, b \sin 	heta) = (5 \cos 	heta, b \sin 	heta)$.The focal distance $S^{\prime}P$ for a point $P(x, y)$ is given by $a + ex$ if $S^{\prime}$ is the left focus $(-ae, 0)$.Thus,$S^{\prime}P = a + ex = 5 + 5(\frac{4}{5}) \cos 	heta = 5 + 4 \cos 	heta$.Given $S^{\prime}P = 7$,we have $5 + 4 \cos 	heta = 7$.$4 \cos 	heta = 2 \Rightarrow \cos 	heta = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

$S^{\prime}$ is the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1, (b < 5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipse. If the distance between the foci of this ellipse is $8$ and $S^{\prime}P = 7$,then $\theta =$

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