What is the vertices 81x^2-1296x+16y^2+64y=-3952 ?

The vertices 81x^2-1296x+16y^2+64y=-3952 is (8,7),(8,-11)

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vertices 81x^2-1296x+16y^2+64y=-3952

vertices

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

(8, 7), (8, - 11)

Solution steps

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

Calculate ellipse properties

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952 :

Ellipse with center

(h, k) = (8, - 2), b = 9, a = 4

(8, - 2 + 9), (8, - 2 - 9)

Refine

(8, 7), (8, - 11)

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

49 x^{2} + 25 y^{2} - 196 x + 250 y - 404 = 0

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

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

What is the vertices 81x^2-1296x+16y^2+64y=-3952 ?
The vertices 81x^2-1296x+16y^2+64y=-3952 is (8,7),(8,-11)