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(A) The given equation is $(y - 1)^2 = 4(x + 1)$.Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get the vertex $(h, k) = (-1, 1)$ an

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$(0, 3)$ and $(0, -1)$

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(A) The given equation is $(y - 1)^2 = 4(x + 1)$.Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get the vertex $(h, k) = (-1, 1)$ and $4a = 4$,so $a = 1$.The latus rectum of a parabola $Y^2 = 4aX$ has endpoints at $(a, 2a)$ and $(a, -2a)$.Here,$X = x - h = x + 1$ and $Y = y - k = y - 1$.Substituting $a = 1$,the endpoints in terms of $(X, Y)$ are $(1, 2)$ and $(1, -2)$.Now,converting back to $(x, y)$ coordinates:For $(1, 2)$: $x + 1 = 1 \implies x = 0$ and $y - 1 = 2 \implies y = 3$.For $(1, -2)$: $x + 1 = 1 \implies x = 0$ and $y - 1 = -2 \implies y = -1$.Thus,the endpoints are $(0, 3)$ and $(0, -1)$.

What are the coordinates of the endpoints of the latus rectum of the parabola $(y - 1)^2 = 4(x + 1)$?

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