What is the foci (x+4)^2=-6(y-3) ?

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

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foci (x+4)^2=-6(y-3)

foci

(x + 4)^{2} = - 6 (y - 3)

(- 4, \frac{3}{2})

Solution steps

Rewrite

(x + 4)^{2} = - 6 (y - 3)

in the standard form:

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

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

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

p

from the center

(- 4, 3)

along the y-axis

(- 4, 3 + p)

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

Refine

(- 4, \frac{3}{2})

(y + 1)^{2} = 2 (x - 5)

y = 5 x^{2} - 12 x + 7

4 x = y^{2} - 4 y

3 x^{2} - 12 x + 2 y + 15 = 0

y^{2} - 3 y