What is the foci x^2+10x=-12y-1 ?

The foci x^2+10x=-12y-1 is (-5,-1)

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foci x^2+10x=-12y-1

foci

x^{2} + 10 x = - 12 y - 1

(- 5, - 1)

Solution steps

Rewrite

x^{2} + 10 x = - 12 y - 1

in the standard form

4 (- 3) (y - 2) = (x - (- 5))^{2}

(h, k) = (- 5, 2), p = - 3

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

p

from the center

(- 5, 2)

along the y-axis

(- 5, 2 + p)

= (- 5, 2 + (- 3))

Refine

(- 5, - 1)

y = (x - 2) (x + 1)

x = 5 (y + 3)^{2} + 6

y = \frac{3 \cdot \frac{1}{2} x ^{2}}{2}

5 x^{2} + 3

(x - 2)^{2} = 4 (y - 2)