What is the foci y=-2x^2-3 ?

The foci y=-2x^2-3 is (0,-25/8)

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foci y=-2x^2-3

foci

y = - 2 x^{2} - 3

(0, - \frac{25}{8})

Solution steps

Rewrite

y = - 2 x^{2} - 3

in the standard form:

4 (- \frac{1}{8}) (y - (- 3)) = (x - 0)^{2}

(h, k) = (0, - 3), p = - \frac{1}{8}

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

p

from the center

(0, - 3)

along the y-axis

(0, - 3 + p)

= (0, - 3 + (- \frac{1}{8}))

Refine

(0, - \frac{25}{8})

(x - 3)^{2} = - 8 (y - 4)

- 2 x^{2} + 6 x - 3

y = 6 x^{2} - 29 x - 5

y = 1 - 3 x - x^{2}

x^{2} - 2 x - 4 y + 13 = 0