What is the vertices 36x^2+16y^2-64y-512=0 ?

The vertices 36x^2+16y^2-64y-512=0 is (0,8),(0,-4)

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vertices 36x^2+16y^2-64y-512=0

vertices

36 x^{2} + 16 y^{2} - 64 y - 512 = 0

(0, 8), (0, - 4)

Solution steps

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

Calculate ellipse properties

36 x^{2} + 16 y^{2} - 64 y - 512 = 0 :

Ellipse with center

(h, k) = (0, 2), b = 6, a = 4

(0, 2 + 6), (0, 2 - 6)

Refine

(0, 8), (0, - 4)

7 x^{2} + 16 y^{2} = 112

9 x^{2} + 49 y^{2} = 441

16 x^{2} + 9 y^{2} - 96 x + 36 y - 36 = 0

\frac{x ^{2}}{7 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

25 x^{2} + 9 y^{2} - 250 x + 72 y + 544 = 0

What is the vertices 36x^2+16y^2-64y-512=0 ?
The vertices 36x^2+16y^2-64y-512=0 is (0,8),(0,-4)