What is the foci 12(x+4)^2+3(y-1)^2=48 ?

The foci 12(x+4)^2+3(y-1)^2=48 is (-4,1+2sqrt(3)),(-4,1-2sqrt(3))

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foci 12(x+4)^2+3(y-1)^2=48

foci

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

(- 4, 1 + 23), (- 4, 1 - 23)

Solution steps

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

Calculate ellipse properties

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48 :

Ellipse with center

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

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

Compute

c :

c = 4^{2} - 2^{2} : 23

(- 4, 1 + 23), (- 4, 1 - 23)

16 x^{2} + 9 y^{2} - 144 = 0

x^{2} + 3 y^{2} = 9

x^{2} + 3 y^{2} = 3

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

9 x^{2} + 25 y^{2} + 18 x - 50 y = 191

What is the foci 12(x+4)^2+3(y-1)^2=48 ?
The foci 12(x+4)^2+3(y-1)^2=48 is (-4,1+2sqrt(3)),(-4,1-2sqrt(3))