What is the foci x^2+5y^2-4x+20y=-19 ?

The foci x^2+5y^2-4x+20y=-19 is (4,-2),(0,-2)

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foci x^2+5y^2-4x+20y=-19

foci

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

(4, - 2), (0, - 2)

Solution steps

(h + c, k), (h - c, k)

Calculate ellipse properties

x^{2} + 5 y^{2} - 4 x + 20 y = - 19 :

Ellipse with center

(h, k) = (2, - 2), a = 5, b = 1

(2 + c, - 2), (2 - c, - 2)

Compute

c :

c = 5^{2} - 1^{2} : 2

(2 + 2, - 2), (2 - 2, - 2)

Refine

(4, - 2), (0, - 2)

\frac{( x - 1 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{9} = 1

x^{2} + \frac{y ^{2}}{9} = 1

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

49 x^{2} + 25 y^{2} - 1225 = 0

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

What is the foci x^2+5y^2-4x+20y=-19 ?
The foci x^2+5y^2-4x+20y=-19 is (4,-2),(0,-2)