What is the foci 16x^2+4y^2-32x+24y-12=0 ?

The foci 16x^2+4y^2-32x+24y-12=0 is (1,-3+2sqrt(3)),(1,-3-2sqrt(3))

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foci 16x^2+4y^2-32x+24y-12=0

foci

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

(1, - 3 + 23), (1, - 3 - 23)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0 :

Ellipse with center

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

(1, - 3 + c), (1, - 3 - c)

Compute

c :

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

(1, - 3 + 23), (1, - 3 - 23)

16 x^{2} + 25 y^{2} - 64 x - 336 = 0

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

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

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

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

What is the foci 16x^2+4y^2-32x+24y-12=0 ?
The foci 16x^2+4y^2-32x+24y-12=0 is (1,-3+2sqrt(3)),(1,-3-2sqrt(3))