What is the foci x=-1/12 (y+10)^2-6 ?

The foci x=-1/12 (y+10)^2-6 is (-9,-10)

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foci x=-1/12 (y+10)^2-6

foci

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

(- 9, - 10)

Solution steps

Rewrite

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

in the standard form:

4 (- 3) (x - (- 6)) = (y - (- 10))^{2}

(h, k) = (- 6, - 10), p = - 3

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

p

from the center

(- 6, - 10)

along the x-axis

(- 6 + p, - 10)

= (- 6 + (- 3), - 10)

Refine

(- 9, - 10)

y = 2 x^{2} - 9

y = 2 x^{2} - 3

y = x^{2} + 8 x + 18

5 x^{2} + 6 x - 4

y = 3 x^{2} - x - 4