What is the foci 2x^2-12x+14-14y=0 ?

The foci 2x^2-12x+14-14y=0 is (3, 41/28)

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foci 2x^2-12x+14-14y=0

foci

2 x^{2} - 12 x + 14 - 14 y = 0

(3, \frac{41}{28})

Solution steps

Rewrite

2 x^{2} - 12 x + 14 - 14 y = 0

in the standard form

4 \cdot \frac{7}{4} (y - (- \frac{2}{7})) = (x - 3)^{2}

(h, k) = (3, - \frac{2}{7}), p = \frac{7}{4}

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

p

from the center

(3, - \frac{2}{7})

along the y-axis

(3, - \frac{2}{7} + p)

= (3, - \frac{2}{7} + \frac{7}{4})

Refine

(3, \frac{41}{28})

x = - 3 y^{2} + 5

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

6 x^{2} - 24 x - 30

x^{2} + y = 4

y = x^{2} - 4 x - 12