What is the vertices 9x^2+4y^2+18x-8y-131=0 ?

The vertices 9x^2+4y^2+18x-8y-131=0 is (-1,7),(-1,-5)

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vertices 9x^2+4y^2+18x-8y-131=0

vertices

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

(- 1, 7), (- 1, - 5)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0 :

Ellipse with center

(h, k) = (- 1, 1), b = 6, a = 4

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

Refine

(- 1, 7), (- 1, - 5)

25 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

\frac{( x + 1 ) ^{2}}{3} + \frac{( y - 1 ) ^{2}}{6} = 4

(x + 3)^{2} + y^{2} = 36

16 x^{2} + 25 y^{2} - 64 x + 150 y - 111 = 0

What is the vertices 9x^2+4y^2+18x-8y-131=0 ?
The vertices 9x^2+4y^2+18x-8y-131=0 is (-1,7),(-1,-5)