What is the foci (x-5)^2=12(y+3) ?

The foci (x-5)^2=12(y+3) is (5,0)

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foci (x-5)^2=12(y+3)

foci

(x - 5)^{2} = 12 (y + 3)

(5, 0)

Solution steps

Rewrite

(x - 5)^{2} = 12 (y + 3)

in the standard form:

4 \cdot 3 (y - (- 3)) = (x - 5)^{2}

(h, k) = (5, - 3), p = 3

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

p

from the center

(5, - 3)

along the y-axis

(5, - 3 + p)

= (5, - 3 + 3)

Refine

(5, 0)

y = - 12 x^{2} - 29 x + 230

\frac{6 x}{7} \frac{4 x}{3}

4 t^{2}

2 x^{2} = - y

y = - x^{2} - 6 x - 5