What is the foci y^2+10y-12x+37=0 ?

The foci y^2+10y-12x+37=0 is (4,-5)

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foci y^2+10y-12x+37=0

foci

y^{2} + 10 y - 12 x + 37 = 0

(4, - 5)

Solution steps

Rewrite

y^{2} + 10 y - 12 x + 37 = 0

in the standard form

4 \cdot 3 (x - 1) = (y - (- 5))^{2}

(h, k) = (1, - 5), p = 3

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

p

from the center

(1, - 5)

along the x-axis

(1 + p, - 5)

= (1 + 3, - 5)

Refine

(4, - 5)

4 x^{2} - 4 x + 3

- 2 x^{2} + 4 x - 6

y^{2} - 4 y - 8 x + 44 = 0

y^{2} + 8 y - 12 x - 20 = 0

- 9 x^{2}