What is the foci 18x^2+y=0 ?

The foci 18x^2+y=0 is (0,-1/72)

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foci 18x^2+y=0

foci

18 x^{2} + y = 0

(0, - \frac{1}{72})

Solution steps

Rewrite

18 x^{2} + y = 0

in the standard form:

4 (- \frac{1}{72}) (y - 0) = (x - 0)^{2}

(h, k) = (0, 0), p = - \frac{1}{72}

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

p

from the center

(0, 0)

along the y-axis

(0, 0 + p)

= (0, 0 + (- \frac{1}{72}))

Refine

(0, - \frac{1}{72})

(x + 1)^{2} = - 3 (y - 6)

y = 3 x^{2} - 11 x + 10

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

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

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