What is the foci y^2+14y+4x+45=0 ?

The foci y^2+14y+4x+45=0 is (0,-7)

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foci y^2+14y+4x+45=0

foci

y^{2} + 14 y + 4 x + 45 = 0

(0, - 7)

Solution steps

Rewrite

y^{2} + 14 y + 4 x + 45 = 0

in the standard form

4 (- 1) (x - 1) = (y - (- 7))^{2}

(h, k) = (1, - 7), p = - 1

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

p

from the center

(1, - 7)

along the x-axis

(1 + p, - 7)

= (1 + (- 1), - 7)

Refine

(0, - 7)

+ (x + 2)^{2} - 4

(x - 4)^{2} = - 8 (y + 2)

- 3 x^{2} - x + 4

y = \frac{( - 6 ( r + 4 r ^{2} ) + 9 r ^{2} + 6 r ) ^{2}}{9 r ^{2}}

y^{2} + 2 y - 4 x + 9 = 0