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

The foci y=(x-3)^2+2 is (3, 9/4)

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foci y=(x-3)^2+2

foci

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

(3, \frac{9}{4})

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{4} (y - 2) = (x - 3)^{2}

(h, k) = (3, 2), p = \frac{1}{4}

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

p

from the center

(3, 2)

along the y-axis

(3, 2 + p)

= (3, 2 + \frac{1}{4})

Refine

(3, \frac{9}{4})

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

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

y = (x - 3)^{2} - 9

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

x^{2} + 5