What is the foci 4x^2+y^2-32x+4y+64=0 ?

The foci 4x^2+y^2-32x+4y+64=0 is (4,-2+sqrt(3)),(4,-2-sqrt(3))

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foci 4x^2+y^2-32x+4y+64=0

foci

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

(4, - 2 + 3), (4, - 2 - 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(4, - 2 + c), (4, - 2 - c)

Compute

c :

c = 2^{2} - 1^{2} : 3

(4, - 2 + 3), (4, - 2 - 3)

\frac{( x ) ^{2}}{144} + \frac{( y ) ^{2}}{169} = 1

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{9} = 1

2 x^{2} + 3 y^{2} + 4 x = 1

\frac{x ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

\frac{( x - 1 ) ^{2}}{25} + \frac{y ^{2}}{16} = 1

What is the foci 4x^2+y^2-32x+4y+64=0 ?
The foci 4x^2+y^2-32x+4y+64=0 is (4,-2+sqrt(3)),(4,-2-sqrt(3))