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

The foci y=-x^2-x-1 is (-1/2 ,-1)

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foci y=-x^2-x-1

foci

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

(- \frac{1}{2}, - 1)

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (- \frac{1}{2}, - \frac{3}{4}), p = - \frac{1}{4}

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

p

from the center

(- \frac{1}{2}, - \frac{3}{4})

along the y-axis

(- \frac{1}{2}, - \frac{3}{4} + p)

= (- \frac{1}{2}, - \frac{3}{4} + (- \frac{1}{4}))

Refine

(- \frac{1}{2}, - 1)

y^{2} - 8 x = 0

- 2 x^{2} + 3 x - 1

(x + 5)^{2}

x^{2} - 2 x + 8 y + 9 = 0

2 x^{2} + 1