What is the foci x=-1/2 (y-4)^2+1/2 ?

The foci x=-1/2 (y-4)^2+1/2 is (0,4)

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foci x=-1/2 (y-4)^2+1/2

foci

x = - \frac{1}{2} (y - 4)^{2} + \frac{1}{2}

(0, 4)

Solution steps

Rewrite

x = - \frac{1}{2} (y - 4)^{2} + \frac{1}{2}

in the standard form:

4 (- \frac{1}{2}) (x - \frac{1}{2}) = (y - 4)^{2}

(h, k) = (\frac{1}{2}, 4), p = - \frac{1}{2}

Parabola is symmetric around the x-axis and so the focus lies a distance

p

from the center

(\frac{1}{2}, 4)

along the x-axis

(\frac{1}{2} + p, 4)

= (\frac{1}{2} + (- \frac{1}{2}), 4)

Refine

(0, 4)

x^{2} + 8 x - 4 y - 36 = 0

4 x^{2} + 8 x + 6

y = (2 x - 1) (5 x - 6)

y = x^{2} - 5 x + 6

x^{2} - x - 56