What is the foci y=(x-3)^2-9 ?

The foci y=(x-3)^2-9 is (3,-35/4)

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foci y=(x-3)^2-9

foci

y = (x - 3)^{2} - 9

(3, - \frac{35}{4})

Solution steps

Rewrite

y = (x - 3)^{2} - 9

in the standard form:

4 \cdot \frac{1}{4} (y - (- 9)) = (x - 3)^{2}

(h, k) = (3, - 9), p = \frac{1}{4}

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

p

from the center

(3, - 9)

along the y-axis

(3, - 9 + p)

= (3, - 9 + \frac{1}{4})

Refine

(3, - \frac{35}{4})

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

x^{2} + 5

y = x^{2} - 64

y = 5 x^{2} + 6 x - 7

2 x^{2} - 12 x + 3