What is the vertices ((x+3)^2)/9+((y-1)^2)/(25)=1 ?

The vertices ((x+3)^2)/9+((y-1)^2)/(25)=1 is (-3,6),(-3,-4)

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vertices ((x+3)^2)/9+((y-1)^2)/(25)=1

vertices

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

(- 3, 6), (- 3, - 4)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1 :

Ellipse with center

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

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

Refine

(- 3, 6), (- 3, - 4)

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

9 x^{2} + 4 y^{2} - 36 x + 40 y + 100 = 0

9 x^{2} + 16 y^{2} + 96 y = 0

49 x^{2} + 25 y^{2} - 1225 = 0

16 x^{2} + 64 y^{2} = 128

What is the vertices ((x+3)^2)/9+((y-1)^2)/(25)=1 ?
The vertices ((x+3)^2)/9+((y-1)^2)/(25)=1 is (-3,6),(-3,-4)