What is the foci (x+1)^2=-3(y-6) ?

The foci (x+1)^2=-3(y-6) is (-1, 21/4)

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foci (x+1)^2=-3(y-6)

foci

(x + 1)^{2} = - 3 (y - 6)

(- 1, \frac{21}{4})

Solution steps

Rewrite

(x + 1)^{2} = - 3 (y - 6)

in the standard form:

4 (- \frac{3}{4}) (y - 6) = (x - (- 1))^{2}

(h, k) = (- 1, 6), p = - \frac{3}{4}

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

p

from the center

(- 1, 6)

along the y-axis

(- 1, 6 + p)

= (- 1, 6 + (- \frac{3}{4}))

Refine

(- 1, \frac{21}{4})

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

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}