What is the foci (y-5)^2-4 ?

The foci (y-5)^2-4 is (-15/4 ,5)

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foci (y-5)^2-4

foci

(y - 5)^{2} - 4

(- \frac{15}{4}, 5)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 4, 5)

along the x-axis

(- 4 + p, 5)

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

Refine

(- \frac{15}{4}, 5)

- x^{2} = 48 y

3 x^{2} + x - 6

- \frac{1}{20} y^{2}

5 x^{2} - 3 x + 6

(x + 1)^{2} = - 12 (y - 2)