What is the foci x=((y-3)^2)/8+5 ?

The foci x=((y-3)^2)/8+5 is (7,3)

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foci x=((y-3)^2)/8+5

foci

x = \frac{( y - 3 ) ^{2}}{8} + 5

(7, 3)

Solution steps

Rewrite

x = \frac{( y - 3 ) ^{2}}{8} + 5

in the standard form:

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

(h, k) = (5, 3), p = 2

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

p

from the center

(5, 3)

along the x-axis

(5 + p, 3)

= (5 + 2, 3)

Refine

(7, 3)

y = - 2 x^{2}

y = + \frac{1}{7} x^{2}

y = 2 x^{2} + x - 15

x = y^{2} + 4 y + 10

(x + 7)^{2} = - 12 (y - 7)