What is the foci 9y^2+12x=0 ?

The foci 9y^2+12x=0 is (-1/3 ,0)

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foci 9y^2+12x=0

foci

9 y^{2} + 12 x = 0

(- \frac{1}{3}, 0)

Solution steps

Rewrite

9 y^{2} + 12 x = 0

in the standard form:

4 (- \frac{1}{3}) (x - 0) = (y - 0)^{2}

(h, k) = (0, 0), p = - \frac{1}{3}

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

p

from the center

(0, 0)

along the x-axis

(0 + p, 0)

= (0 + (- \frac{1}{3}), 0)

Refine

(- \frac{1}{3}, 0)

8 y = x^{2}

y = 2 x^{2} - 7 x - 3

12 x^{2} - 72 x + 77

y^{2} - 10 x + 4 y - 16 = 0

x^{2} - 4 x - 5