What is the foci y=(x+5)^2 ?

The foci y=(x+5)^2 is (-5, 1/4)

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foci y=(x+5)^2

foci

y = (x + 5)^{2}

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

Solution steps

Rewrite

y = (x + 5)^{2}

in the standard form:

4 \cdot \frac{1}{4} (y - 0) = (x - (- 5))^{2}

(h, k) = (- 5, 0), p = \frac{1}{4}

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

p

from the center

(- 5, 0)

along the y-axis

(- 5, 0 + p)

= (- 5, 0 + \frac{1}{4})

Refine

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

x^{2} + 2 x + 6

y = 2 x^{2} - 4 x + 15

x^{2} + 2 x - 2

2 y^{2} = 3 x

(y + 5)^{2} = - 12 (x + 1)