What is the foci y=-4x^2+1 ?

The foci y=-4x^2+1 is (0, 15/16)

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foci y=-4x^2+1

foci

y = - 4 x^{2} + 1

(0, \frac{15}{16})

Solution steps

Rewrite

y = - 4 x^{2} + 1

in the standard form:

4 (- \frac{1}{16}) (y - 1) = (x - 0)^{2}

(h, k) = (0, 1), p = - \frac{1}{16}

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

p

from the center

(0, 1)

along the y-axis

(0, 1 + p)

= (0, 1 + (- \frac{1}{16}))

Refine

(0, \frac{15}{16})

x^{2} - 20 y - 2 x - 39 = 0

y^{2} + x + y = 0

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

3 x^{2} - 16 x + 5

y^{2} - 12 y + 6 x + 36 = 0