What is the vertices 25x^2+9y^2+50x-108y+124=0 ?

The vertices 25x^2+9y^2+50x-108y+124=0 is (-1,11),(-1,1)

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vertices 25x^2+9y^2+50x-108y+124=0

vertices

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

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

Solution steps

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

Calculate ellipse properties

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0 :

Ellipse with center

(h, k) = (- 1, 6), b = 5, a = 3

(- 1, 6 + 5), (- 1, 6 - 5)

Refine

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

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

3 x^{2} + 4 y^{2} = 12

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{4} = 1

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

What is the vertices 25x^2+9y^2+50x-108y+124=0 ?
The vertices 25x^2+9y^2+50x-108y+124=0 is (-1,11),(-1,1)