What is the vertices x^2+10x+9y^2+16=0 ?

The vertices x^2+10x+9y^2+16=0 is (-2,0),(-8,0)

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vertices x^2+10x+9y^2+16=0

vertices

x^{2} + 10 x + 9 y^{2} + 16 = 0

(- 2, 0), (- 8, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

x^{2} + 10 x + 9 y^{2} + 16 = 0 :

Ellipse with center

(h, k) = (- 5, 0), a = 3, b = 1

(- 5 + 3, 0), (- 5 - 3, 0)

Refine

(- 2, 0), (- 8, 0)

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

16 x^{2} - 32 x + 9 y^{2} = 128

What is the vertices x^2+10x+9y^2+16=0 ?
The vertices x^2+10x+9y^2+16=0 is (-2,0),(-8,0)