What is the foci (x+7)^2=5(y+3) ?

The foci (x+7)^2=5(y+3) is (-7,-7/4)

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foci (x+7)^2=5(y+3)

foci

(x + 7)^{2} = 5 (y + 3)

(- 7, - \frac{7}{4})

Solution steps

Rewrite

(x + 7)^{2} = 5 (y + 3)

in the standard form:

4 \cdot \frac{5}{4} (y - (- 3)) = (x - (- 7))^{2}

(h, k) = (- 7, - 3), p = \frac{5}{4}

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

p

from the center

(- 7, - 3)

along the y-axis

(- 7, - 3 + p)

= (- 7, - 3 + \frac{5}{4})

Refine

(- 7, - \frac{7}{4})

x^{2} - 8 y = 0

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

- \frac{x ^{2}}{8} + \frac{x}{4} + \frac{23}{8}

3 x^{2} + 8

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