What is the foci y=2(x-5)^2-8 ?

The foci y=2(x-5)^2-8 is (5,-63/8)

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foci y=2(x-5)^2-8

foci

y = 2 (x - 5)^{2} - 8

(5, - \frac{63}{8})

Solution steps

Rewrite

y = 2 (x - 5)^{2} - 8

in the standard form:

4 \cdot \frac{1}{8} (y - (- 8)) = (x - 5)^{2}

(h, k) = (5, - 8), p = \frac{1}{8}

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

p

from the center

(5, - 8)

along the y-axis

(5, - 8 + p)

= (5, - 8 + \frac{1}{8})

Refine

(5, - \frac{63}{8})

y = + 2 x^{2} + 2

\frac{1}{2} (x - 3)^{2} + 4

- x^{2} + 3 x - 5

y = - \frac{1}{3} x^{2}

x^{2} = 8 y + 2 x - 33