What is the foci (x-5)^2-2(y+2)=0 ?

The foci (x-5)^2-2(y+2)=0 is (5,-3/2)

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foci (x-5)^2-2(y+2)=0

foci

(x - 5)^{2} - 2 (y + 2) = 0

(5, - \frac{3}{2})

Solution steps

Rewrite

(x - 5)^{2} - 2 (y + 2) = 0

in the standard form:

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

(h, k) = (5, - 2), p = \frac{1}{2}

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

p

from the center

(5, - 2)

along the y-axis

(5, - 2 + p)

= (5, - 2 + \frac{1}{2})

Refine

(5, - \frac{3}{2})

(x + 5)^{2} = - 12 (y - 3)

y = (x - 4) (x - 7)

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

y^{2} + 4 x = 0

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