What is the foci y=3x^2+6 ?

The foci y=3x^2+6 is (0, 73/12)

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foci y=3x^2+6

foci

y = 3 x^{2} + 6

(0, \frac{73}{12})

Solution steps

Rewrite

y = 3 x^{2} + 6

in the standard form:

4 \cdot \frac{1}{12} (y - 6) = (x - 0)^{2}

(h, k) = (0, 6), p = \frac{1}{12}

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

p

from the center

(0, 6)

along the y-axis

(0, 6 + p)

= (0, 6 + \frac{1}{12})

Refine

(0, \frac{73}{12})

y = 2 x^{2} - x

x = - \frac{1}{12} (y + 10)^{2} - 6

y = 2 x^{2} - 9

y = 2 x^{2} - 3

y = x^{2} + 8 x + 18