What is the vertices 9x^2+4y^2+72x-40y+208=0 ?

The vertices 9x^2+4y^2+72x-40y+208=0 is (-4,8),(-4,2)

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vertices 9x^2+4y^2+72x-40y+208=0

vertices

9 x^{2} + 4 y^{2} + 72 x - 40 y + 208 = 0

(- 4, 8), (- 4, 2)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 4 y^{2} + 72 x - 40 y + 208 = 0 :

Ellipse with center

(h, k) = (- 4, 5), b = 3, a = 2

(- 4, 5 + 3), (- 4, 5 - 3)

Refine

(- 4, 8), (- 4, 2)

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

3 x^{2} + 7 y^{2} - 12 x - 28 y + 19 = 0

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

9 x^{2} + 16 y^{2} = 576

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

What is the vertices 9x^2+4y^2+72x-40y+208=0 ?
The vertices 9x^2+4y^2+72x-40y+208=0 is (-4,8),(-4,2)