Let L(ae, b^2/a) be the end of the latus rect | vedclass

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(B) The coordinates of the focus $S$ are $(ae, 0)$ and the end of the latus rectum $L$ in the first quadrant is $(ae, b^2/a)$.Given $S(8, y_1)$,we hav

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$4(\sqrt{17}-1)$

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(B) The coordinates of the focus $S$ are $(ae, 0)$ and the end of the latus rectum $L$ in the first quadrant is $(ae, b^2/a)$.Given $S(8, y_1)$,we have $ae = 8$. Since $y_1$ is the $y$-coordinate of the focus,$y_1 = 0$.Given $L(x_1, 4)$,we have $b^2/a = 4$,which implies $b^2 = 4a$.For a hyperbola,$c^2 = a^2 + b^2$,where $c = ae = 8$.Substituting $c = 8$ and $b^2 = 4a$ into the equation $c^2 = a^2 + b^2$:$64 = a^2 + 4a$$a^2 + 4a - 64 = 0$Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$a = \frac{-4 \pm \sqrt{16 - 4(1)(-64)}}{2} = \frac{-4 \pm \sqrt{16 + 256}}{2} = \frac{-4 \pm \sqrt{272}}{2} = \frac{-4 \pm 4\sqrt{17}}{2} = -2 \pm 2\sqrt{17}$.Since $a > 0$,we take $a = 2\sqrt{17} - 2 = 2(\sqrt{17} - 1)$.The length of the transverse axis is $2a = 2 	imes 2(\sqrt{17} - 1) = 4(\sqrt{17} - 1)$.

Let $L(ae, b^2/a)$ be the end of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ lying in the first quadrant,and let $S(ae, 0)$ be the focus of the given hyperbola. Given $L$ is $(x_1, 4)$ and $S$ is $(8, y_1)$,find the length of its transverse axis.

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