What is the foci y=2x^2+28x+105 ?

The foci y=2x^2+28x+105 is (-7, 57/8)

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foci y=2x^2+28x+105

foci

y = 2 x^{2} + 28 x + 105

(- 7, \frac{57}{8})

Solution steps

Rewrite

y = 2 x^{2} + 28 x + 105

in the standard form

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

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

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

p

from the center

(- 7, 7)

along the y-axis

(- 7, 7 + p)

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

Refine

(- 7, \frac{57}{8})

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

x^{2} - 4 x + 3

(x - 4)^{2} = - 12 (y - 3)

x^{2} - 2 x - 24

X^{2} - 5 X - 24