What is the vertices ((x+2)^2)/9+((y-1)^2)/(25)=1 ?

The vertices ((x+2)^2)/9+((y-1)^2)/(25)=1 is (-2,6),(-2,-4)

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vertices ((x+2)^2)/9+((y-1)^2)/(25)=1

vertices

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

(- 2, 6), (- 2, - 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 1), b = 5, a = 3

(- 2, 1 + 5), (- 2, 1 - 5)

Refine

(- 2, 6), (- 2, - 4)

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

9 x^{2} + 16 y^{2} - 144 = 0

What is the vertices ((x+2)^2)/9+((y-1)^2)/(25)=1 ?
The vertices ((x+2)^2)/9+((y-1)^2)/(25)=1 is (-2,6),(-2,-4)