What is the vertices ((x+5)^2)/9+(y-1)^2=1 ?

The vertices ((x+5)^2)/9+(y-1)^2=1 is (-2,1),(-8,1)

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vertices ((x+5)^2)/9+(y-1)^2=1

vertices

\frac{( x + 5 ) ^{2}}{9} + (y - 1)^{2} = 1

(- 2, 1), (- 8, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x + 5 ) ^{2}}{9} + (y - 1)^{2} = 1 :

Ellipse with center

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

(- 5 + 3, 1), (- 5 - 3, 1)

Refine

(- 2, 1), (- 8, 1)

8 x^{2} + 2 y^{2} + 16 x = 0

9 X^{2} + 4 Y^{2} = 36

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

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

2 x^{2} + y^{2} = 50

What is the vertices ((x+5)^2)/9+(y-1)^2=1 ?
The vertices ((x+5)^2)/9+(y-1)^2=1 is (-2,1),(-8,1)