What is the foci y=(x+3)^2 ?

The foci y=(x+3)^2 is (-3, 1/4)

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foci y=(x+3)^2

foci

y = (x + 3)^{2}

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

Solution steps

Rewrite

y = (x + 3)^{2}

in the standard form:

4 \cdot \frac{1}{4} (y - 0) = (x - (- 3))^{2}

(h, k) = (- 3, 0), p = \frac{1}{4}

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

p

from the center

(- 3, 0)

along the y-axis

(- 3, 0 + p)

= (- 3, 0 + \frac{1}{4})

Refine

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

(x + 2)^{2} = - 8 (y - 1)

y^{2} = 5 x + 3

(x + 3)^{2} = - 2 (y - 1)

y = (x + \frac{1}{4}) (x - 5)

y^{2} - 16 x = 0