What is the foci x=(y+4)^2 ?

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

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foci x=(y+4)^2

foci

x = (y + 4)^{2}

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

Solution steps

Rewrite

x = (y + 4)^{2}

in the standard form:

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

(h, k) = (0, - 4), p = \frac{1}{4}

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

p

from the center

(0, - 4)

along the x-axis

(0 + p, - 4)

= (0 + \frac{1}{4}, - 4)

Refine

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

y = 6 x^{2} - 1

1.5 + 17.67 t - 4.905 t^{2}

y = (3 x - 4) (4 x - 3) - 2 x

x^{2} - 4 x - y + 3 = 0

- 2 x^{2} + 24 x + y - 70 = 0