What is the foci y=4x+4x-2+x^2-3 ?

The foci y=4x+4x-2+x^2-3 is (-4,-83/4)

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foci y=4x+4x-2+x^2-3

foci

y = 4 x + 4 x - 2 + x^{2} - 3

(- 4, - \frac{83}{4})

Solution steps

Rewrite

y = 4 x + 4 x - 2 + x^{2} - 3

in the standard form

4 \cdot \frac{1}{4} (y - (- 21)) = (x - (- 4))^{2}

(h, k) = (- 4, - 21), p = \frac{1}{4}

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

p

from the center

(- 4, - 21)

along the y-axis

(- 4, - 21 + p)

= (- 4, - 21 + \frac{1}{4})

Refine

(- 4, - \frac{83}{4})

y = (x - 5)^{2} - 9

25 x^{2} - 10 x - 200 y - 119 = 0

2 π r^{2} + 16 π r

x^{2} - 4 x - 8 y - 12 = 0

(x - 6)^{2} = - 4 y