What is the foci 41x^2+16y^2+246x-192y+289=0 ?

The foci 41x^2+16y^2+246x-192y+289=0 is (-3,11),(-3,1)

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foci 41x^2+16y^2+246x-192y+289=0

foci

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

(- 3, 11), (- 3, 1)

Solution steps

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

Calculate ellipse properties

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0 :

Ellipse with center

(h, k) = (- 3, 6), b = 41, a = 4

(- 3, 6 + c), (- 3, 6 - c)

Compute

c :

c = 41^{2} - 4^{2} : 5

(- 3, 6 + 5), (- 3, 6 - 5)

Refine

(- 3, 11), (- 3, 1)

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{16} = 1

4 x^{2} + 25 y^{2} = 100

\frac{y ^{2}}{25} + \frac{x ^{2}}{9} = 1

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

What is the foci 41x^2+16y^2+246x-192y+289=0 ?
The foci 41x^2+16y^2+246x-192y+289=0 is (-3,11),(-3,1)