What is the foci y=3x^2-11x+10 ?

The foci y=3x^2-11x+10 is (11/6 ,0)

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foci y=3x^2-11x+10

foci

y = 3 x^{2} - 11 x + 10

(\frac{11}{6}, 0)

Solution steps

Rewrite

y = 3 x^{2} - 11 x + 10

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{1}{12})) = (x - \frac{11}{6})^{2}

(h, k) = (\frac{11}{6}, - \frac{1}{12}), p = \frac{1}{12}

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

p

from the center

(\frac{11}{6}, - \frac{1}{12})

along the y-axis

(\frac{11}{6}, - \frac{1}{12} + p)

= (\frac{11}{6}, - \frac{1}{12} + \frac{1}{12})

Refine

(\frac{11}{6}, 0)

x^{2} + 10 x = - 12 y - 1

y = (x - 2) (x + 1)

x = 5 (y + 3)^{2} + 6

y = \frac{3 \cdot \frac{1}{2} x ^{2}}{2}

5 x^{2} + 3