Let the foci of a hyperbola be (1, 14) and (1 | vedclass

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(C) The foci are $F_1 = (1, 14)$ and $F_2 = (1, -12)$. The center of the hyperbola is the midpoint of the foci: $(\frac{1+1}{2}, \frac{14-12}{2}) = (1

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$\frac{288}{5}$

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(C) The foci are $F_1 = (1, 14)$ and $F_2 = (1, -12)$. The center of the hyperbola is the midpoint of the foci: $(\frac{1+1}{2}, \frac{14-12}{2}) = (1, 1)$.The distance between the foci is $2ae = \sqrt{(1-1)^2 + (14 - (-12))^2} = \sqrt{0^2 + 26^2} = 26$,so $ae = 13$.The hyperbola passes through $(1, 6)$. Since the transverse axis is vertical (along $x=1$),the distance from a point $P$ on the hyperbola to the foci is $|PF_1 - PF_2| = 2a$.$PF_1 = \sqrt{(1-1)^2 + (14-6)^2} = 8$.$PF_2 = \sqrt{(1-1)^2 + (6 - (-12))^2} = 18$.$2a = |8 - 18| = 10$,so $a = 5$.Using $ae = 13$,we get $5e = 13$,so $e = \frac{13}{5}$.For a hyperbola,$b^2 = a^2(e^2 - 1) = a^2e^2 - a^2 = (ae)^2 - a^2 = 13^2 - 5^2 = 169 - 25 = 144$.Thus,$b^2 = 144$,which means $b = 12$.The length of the latus-rectum is $\frac{2b^2}{a} = \frac{2 	imes 144}{5} = \frac{288}{5}$.

Let the foci of a hyperbola be $(1, 14)$ and $(1, -12)$. If it passes through the point $(1, 6)$,then the length of its latus-rectum is:

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