What is the foci y=+(x-7)^2+8 ?

The foci y=+(x-7)^2+8 is (7, 33/4)

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foci y=+(x-7)^2+8

foci

y = + (x - 7)^{2} + 8

(7, \frac{33}{4})

Solution steps

Rewrite

y = (x - 7)^{2} + 8

in the standard form:

4 \cdot \frac{1}{4} (y - 8) = (x - 7)^{2}

(h, k) = (7, 8), p = \frac{1}{4}

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

p

from the center

(7, 8)

along the y-axis

(7, 8 + p)

= (7, 8 + \frac{1}{4})

Refine

(7, \frac{33}{4})

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

(2 x - 3)^{2}

\frac{x ^{2}}{6}

y = x^{2} + 12 x + 26

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