What is the foci 4(x+2)^2+16(y-2)^2=64 ?

The foci 4(x+2)^2+16(y-2)^2=64 is (-2+2sqrt(3),2),(-2-2sqrt(3),2)

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foci 4(x+2)^2+16(y-2)^2=64

foci

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

(- 2 + 23, 2), (- 2 - 23, 2)

Solution steps

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

Calculate ellipse properties

4 (x + 2)^{2} + 16 (y - 2)^{2} = 64 :

Ellipse with center

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

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

Compute

c :

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

(- 2 + 23, 2), (- 2 - 23, 2)

\frac{( x - 7 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{25} = 1

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{1} = 1

\frac{( x - 1 ) ^{2} + ( y - 3 ) ^{2}}{2} = 1

\frac{x ^{2}}{289} + \frac{y ^{2}}{225} = 1

4 x^{2} + y^{2} - 32 x + 4 y + 64 = 0

What is the foci 4(x+2)^2+16(y-2)^2=64 ?
The foci 4(x+2)^2+16(y-2)^2=64 is (-2+2sqrt(3),2),(-2-2sqrt(3),2)