What is the foci y=p(x-h)^2+k ?

The foci y=p(x-h)^2+k is (h,(4pk+1)/(4p))

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foci y=p(x-h)^2+k

foci

y = p (x - h)^{2} + k

(h, \frac{4 p k + 1}{4 p})

Solution steps

Rewrite

y = p (x - h)^{2} + k

in the standard form:

4 \cdot \frac{1}{4 p} (y - k) = (x - h)^{2}

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

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

p

from the center

(h, k)

along the y-axis

(h, k + p)

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

Refine

(h, \frac{4 p k + 1}{4 p})

(x - 3)^{2} = - 12 y

y = 3 x^{2} + 6

y = 2 x^{2} - x

x = - \frac{1}{12} (y + 10)^{2} - 6

y = 2 x^{2} - 9