What is the foci y=x^2+10x+25 ?

The foci y=x^2+10x+25 is (-5, 1/4)

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foci y=x^2+10x+25

foci

y = x^{2} + 10 x + 25

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

Solution steps

Rewrite

y = x^{2} + 10 x + 25

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} = 8 (y - 3)

x^{2} + 2 x - 12 y + 37 = 0

x^{2} + 4 x + 6 y - 2 = 0

(x + 4)^{2} = - 8 (y - 3)

x^{2} - 2 x + 37