What is the foci x=+(y+1)^2-9 ?

The foci x=+(y+1)^2-9 is (-35/4 ,-1)

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foci x=+(y+1)^2-9

foci

x = + (y + 1)^{2} - 9

(- \frac{35}{4}, - 1)

Solution steps

Rewrite

x = (y + 1)^{2} - 9

in the standard form:

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

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

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

p

from the center

(- 9, - 1)

along the x-axis

(- 9 + p, - 1)

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

Refine

(- \frac{35}{4}, - 1)

- 3 x^{2} + 8 x + 896

x^{2} + 36

y = \frac{40 x ^{3}}{3 x}

x^{2} + 10 x - 2 y + 21 = 0

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