What is the foci y=x^2-4x-12 ?

The foci y=x^2-4x-12 is (2,-63/4)

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foci y=x^2-4x-12

foci

y = x^{2} - 4 x - 12

(2, - \frac{63}{4})

Solution steps

Rewrite

y = x^{2} - 4 x - 12

in the standard form

4 \cdot \frac{1}{4} (y - (- 16)) = (x - 2)^{2}

(h, k) = (2, - 16), p = \frac{1}{4}

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

p

from the center

(2, - 16)

along the y-axis

(2, - 16 + p)

= (2, - 16 + \frac{1}{4})

Refine

(2, - \frac{63}{4})

x^{2} = - y

(y - 2)^{2} = (x - 1)^{2} + (y + 4)^{2}

x^{(2)} - 6 x + 5 y = - 34

x (2 x + 3)

6 x^{2} - 49