What is the foci 6y=x^2-8x+14 ?

The foci 6y=x^2-8x+14 is (4, 7/6)

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foci 6y=x^2-8x+14

foci

6 y = x^{2} - 8 x + 14

(4, \frac{7}{6})

Solution steps

Rewrite

6 y = x^{2} - 8 x + 14

in the standard form

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

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

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

p

from the center

(4, - \frac{1}{3})

along the y-axis

(4, - \frac{1}{3} + p)

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

Refine

(4, \frac{7}{6})

(x - 1)^{2} = - 8 (y 2)

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

- 4 x^{2}

y = 2 x^{2} - 4 x

- 3 x^{2} + 3 x - 13