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

The foci y=(x-2)^2-4 is (2,-15/4)

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foci y=(x-2)^2-4

foci

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

(2, - \frac{15}{4})

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(2, - 4)

along the y-axis

(2, - 4 + p)

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

Refine

(2, - \frac{15}{4})

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

y = x^{2} + 7 x - 1

+ 3 y^{2} = 24 x

y = \frac{18 x ^{6}}{27 x ^{4}}

y^{2} - 2 y - 8 x - 39 = 0