What is the foci (x+3)+(y-2)^2=0 ?

The foci (x+3)+(y-2)^2=0 is (-13/4 ,2)

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foci (x+3)+(y-2)^2=0

foci

(x + 3) + (y - 2)^{2} = 0

(- \frac{13}{4}, 2)

Solution steps

Rewrite

x + 3 + (y - 2)^{2} = 0

in the standard form:

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

(h, k) = (- 3, 2), p = - \frac{1}{4}

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

p

from the center

(- 3, 2)

along the x-axis

(- 3 + p, 2)

= (- 3 + (- \frac{1}{4}), 2)

Refine

(- \frac{13}{4}, 2)

x = 2 y - y^{2}

- \frac{( x + 3 ) ^{2}}{20} + 1

y = 3 x^{2} - 15 x + 6

- 12 y + 10 = x^{2} - 4 x + 14

(x + 4)^{2} = - 6 (y - 3)