What is the foci y=2x^2-4x+15 ?

The foci y=2x^2-4x+15 is (1, 105/8)

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foci y=2x^2-4x+15

foci

y = 2 x^{2} - 4 x + 15

(1, \frac{105}{8})

Solution steps

Rewrite

y = 2 x^{2} - 4 x + 15

in the standard form

4 \cdot \frac{1}{8} (y - 13) = (x - 1)^{2}

(h, k) = (1, 13), p = \frac{1}{8}

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

p

from the center

(1, 13)

along the y-axis

(1, 13 + p)

= (1, 13 + \frac{1}{8})

Refine

(1, \frac{105}{8})

x^{2} + 2 x - 2

2 y^{2} = 3 x

(y + 5)^{2} = - 12 (x + 1)

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

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