What is the vertices x^2+y^2+8x=32 ?

The vertices x^2+y^2+8x=32 is (-4,4sqrt(3)),(-4,-4sqrt(3))

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vertices x^2+y^2+8x=32

vertices

x^{2} + y^{2} + 8 x = 32

(- 4, 43), (- 4, - 43)

Solution steps

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

Calculate ellipse properties

x^{2} + y^{2} + 8 x = 32 :

Ellipse with center

(h, k) = (- 4, 0), b = 43, a = 43

(- 4, 0 + 43), (- 4, 0 - 43)

Refine

(- 4, 43), (- 4, - 43)

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

25 x^{2} + 81 y^{2} = 2025

18 x^{2} + 4 y^{2} - 72 x + 4 y + 37 = 0

4 - 9 x^{2} - 2 y^{2} = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

What is the vertices x^2+y^2+8x=32 ?
The vertices x^2+y^2+8x=32 is (-4,4sqrt(3)),(-4,-4sqrt(3))