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

The vertices 41x^2+16y^2+246x-192y+289=0 is (-3,6+sqrt(41)),(-3,6-sqrt(41))

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

vertices

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

(- 3, 6 + 41), (- 3, 6 - 41)

Solution steps

(h, k + b), (h, k - b)

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 + 41), (- 3, 6 - 41)

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

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

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

9 x^{2} - 18 x + 4 y^{2} = 27

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

What is the vertices 41x^2+16y^2+246x-192y+289=0 ?
The vertices 41x^2+16y^2+246x-192y+289=0 is (-3,6+sqrt(41)),(-3,6-sqrt(41))