If a parabola passes through the points (-2, | vedclass

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(C) The equation of a parabola with a horizontal axis is given by $(y - k)^2 = 4a(x - h)$.Substituting the given points $(-2, 1)$,$(1, 2)$,and $(-1, 3

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$\frac{2}{5}$

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(C) The equation of a parabola with a horizontal axis is given by $(y - k)^2 = 4a(x - h)$.Substituting the given points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$ into the equation:For $(-2, 1)$: $(1 - k)^2 = 4a(-2 - h) \implies 1 - 2k + k^2 = -8a - 4ah$  $(i)$For $(1, 2)$: $(2 - k)^2 = 4a(1 - h) \implies 4 - 4k + k^2 = 4a - 4ah$  (ii)For $(-1, 3)$: $(3 - k)^2 = 4a(-1 - h) \implies 9 - 6k + k^2 = -4a - 4ah$  (iii)Subtracting (ii) from $(i)$: $(1 - 2k + k^2) - (4 - 4k + k^2) = -8a - 4ah - (4a - 4ah) \implies -3 + 2k = -12a \implies 2k = 3 - 12a$  (iv)Subtracting (iii) from (ii): $(4 - 4k + k^2) - (9 - 6k + k^2) = 4a - 4ah - (-4a - 4ah) \implies -5 + 2k = 8a$  $(v)$Substituting (iv) into $(v)$: $(3 - 12a) - 5 = 8a \implies -2 = 20a \implies a = -\frac{1}{10}$.The length of the latus rectum is $|4a| = |4 	imes -\frac{1}{10}| = |-\frac{4}{10}| = \frac{2}{5}$.

If a parabola passes through the points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$ and has a horizontal axis,then the length of the latus rectum of that parabola is:

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