What is the foci 1/2 (x-3)^2+4 ?

The foci 1/2 (x-3)^2+4 is (3, 9/2)

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foci 1/2 (x-3)^2+4

foci

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

(3, \frac{9}{2})

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(3, 4)

along the y-axis

(3, 4 + p)

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

Refine

(3, \frac{9}{2})

- x^{2} + 3 x - 5

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

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

(x - 4)^{2} = 20 (y + 5)

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