What is the vertices ((x+1)^2)/(16)+((y+1)^2)/(25)=1 ?

The vertices ((x+1)^2)/(16)+((y+1)^2)/(25)=1 is (-1,4),(-1,-6)

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vertices ((x+1)^2)/(16)+((y+1)^2)/(25)=1

vertices

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

(- 1, 4), (- 1, - 6)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, - 1), b = 5, a = 4

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

Refine

(- 1, 4), (- 1, - 6)

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

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

5 x^{2} + 2 y^{2} = 5

x^{2} + 3 y^{2} + 6 x - 6 y + 9 = 0

81 x^{2} + 16 y^{2} = 1296

What is the vertices ((x+1)^2)/(16)+((y+1)^2)/(25)=1 ?
The vertices ((x+1)^2)/(16)+((y+1)^2)/(25)=1 is (-1,4),(-1,-6)