What is the foci ((x+4)^2)/9+((y-1)^2)/4 =1 ?

The foci ((x+4)^2)/9+((y-1)^2)/4 =1 is (-4+sqrt(5),1),(-4-sqrt(5),1)

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foci ((x+4)^2)/9+((y-1)^2)/4 =1

foci

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

(- 4 + 5, 1), (- 4 - 5, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 4, 1), a = 3, b = 2

(- 4 + c, 1), (- 4 - c, 1)

Compute

c :

c = 3^{2} - 2^{2} : 5

(- 4 + 5, 1), (- 4 - 5, 1)

\frac{x ^{2}}{27} + \frac{y ^{2}}{36} = 1

x^{2} + 4 x + 2 y^{2} + 6 y + 1 = 0

25 x^{2} + 81 y^{2} = 2025

x^{2} + 9 y^{2} - 14 x + 36 y + 49 = 0

4 x^{2} - 24 x + \frac{y ^{2}}{16} + \frac{y}{8} + \frac{573}{16} = 0

What is the foci ((x+4)^2)/9+((y-1)^2)/4 =1 ?
The foci ((x+4)^2)/9+((y-1)^2)/4 =1 is (-4+sqrt(5),1),(-4-sqrt(5),1)