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(B) The equation of a parabola with a horizontal axis is $x = ay^2 + by + c$.Substituting the given points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$:$1) -2 = a

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$\frac{21}{10}$

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(B) The equation of a parabola with a horizontal axis is $x = ay^2 + by + c$.Substituting the given points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$:$1) -2 = a + b + c$$2) 1 = 4a + 2b + c$$3) -1 = 9a + 3b + c$Subtracting $(1)$ from $(2)$: $3a + b = 3$ $(4)$Subtracting $(2)$ from $(3)$: $5a + b = -2$ $(5)$Subtracting $(4)$ from $(5)$: $2a = -5 \Rightarrow a = -\frac{5}{2}$.Substituting $a$ into $(4)$: $3(-\frac{5}{2}) + b = 3 \Rightarrow b = 3 + \frac{15}{2} = \frac{21}{2}$.Substituting $a$ and $b$ into $(1)$: $-\frac{5}{2} + \frac{21}{2} + c = -2$ $\Rightarrow 8 + c = -2$ $\Rightarrow c = -10$.The parabola is $x = -\frac{5}{2}y^2 + \frac{21}{2}y - 10$.Completing the square for $y$: $x = -\frac{5}{2}(y^2 - \frac{21}{5}y) - 10 = -\frac{5}{2}(y - \frac{21}{10})^2 + \frac{5}{2}(\frac{441}{100}) - 10 = -\frac{5}{2}(y - \frac{21}{10})^2 + \frac{441}{40} - \frac{400}{40} = -\frac{5}{2}(y - \frac{21}{10})^2 + \frac{41}{40}$.Rearranging: $(y - \frac{21}{10})^2 = -\frac{2}{5}(x - \frac{41}{40})$.This is of the form $(y - k)^2 = 4A(x - h)$,where $k = \frac{21}{10}$.The $y$-coordinate of the focus is $k = \frac{21}{10}$.

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

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