What is the foci y=+2x^2+2 ?

The foci y=+2x^2+2 is (0, 17/8)

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foci y=+2x^2+2

foci

y = + 2 x^{2} + 2

(0, \frac{17}{8})

Solution steps

Rewrite

y = 2 x^{2} + 2

in the standard form:

4 \cdot \frac{1}{8} (y - 2) = (x - 0)^{2}

(h, k) = (0, 2), p = \frac{1}{8}

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

p

from the center

(0, 2)

along the y-axis

(0, 2 + p)

= (0, 2 + \frac{1}{8})

Refine

(0, \frac{17}{8})

\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

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