What is the foci y=(x-5)^2-9 ?

The foci y=(x-5)^2-9 is (5,-35/4)

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foci y=(x-5)^2-9

foci

y = (x - 5)^{2} - 9

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

Solution steps

Rewrite

y = (x - 5)^{2} - 9

in the standard form:

4 \cdot \frac{1}{4} (y - (- 9)) = (x - 5)^{2}

(h, k) = (5, - 9), p = \frac{1}{4}

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

p

from the center

(5, - 9)

along the y-axis

(5, - 9 + p)

= (5, - 9 + \frac{1}{4})

Refine

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

25 x^{2} - 10 x - 200 y - 119 = 0

2 π r^{2} + 16 π r

x^{2} - 4 x - 8 y - 12 = 0

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

4 t^{2} + 1