What is the foci y=-3x^2+30x-71 ?

The foci y=-3x^2+30x-71 is (5, 47/12)

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foci y=-3x^2+30x-71

foci

y = - 3 x^{2} + 30 x - 71

(5, \frac{47}{12})

Solution steps

Rewrite

y = - 3 x^{2} + 30 x - 71

in the standard form

4 (- \frac{1}{12}) (y - 4) = (x - 5)^{2}

(h, k) = (5, 4), p = - \frac{1}{12}

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

p

from the center

(5, 4)

along the y-axis

(5, 4 + p)

= (5, 4 + (- \frac{1}{12}))

Refine

(5, \frac{47}{12})

(x - 3)^{2}

y^{2} + 10 y - 12 x + 37 = 0

4 x^{2} - 4 x + 3

- 2 x^{2} + 4 x - 6

y^{2} - 4 y - 8 x + 44 = 0