An ellipse has its center at (1, -2), one foc | vedclass

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(B) Given center $C = (1, -2)$, focus $F_1 = (3, -2)$, and vertex $A_1 = (5, -2)$.Since the $y$-coordinates are the same, the major axis is horizontal

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$6$

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(B) Given center $C = (1, -2)$, focus $F_1 = (3, -2)$, and vertex $A_1 = (5, -2)$.Since the $y$-coordinates are the same, the major axis is horizontal.The distance from the center to the vertex is $a = |5 - 1| = 4$.The distance from the center to the focus is $ae = |3 - 1| = 2$.Thus, $e = \frac{2}{4} = \frac{1}{2}$.Using the relation $b^2 = a^2(1 - e^2)$, we get $b^2 = 4^2(1 - (\frac{1}{2})^2) = 16(1 - \frac{1}{4}) = 16(\frac{3}{4}) = 12$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 	imes 12}{4} = \frac{24}{4} = 6$.

An ellipse has its center at $(1, -2)$, one focus at $(3, -2)$, and one vertex at $(5, -2)$. Then the length of its latus rectum is :

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