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

The vertices ((x+2)^2)/(16)+((y+2)^2)/(25)=1 is (-2,3),(-2,-7)

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

vertices

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

(- 2, 3), (- 2, - 7)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, - 2), b = 5, a = 4

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

Refine

(- 2, 3), (- 2, - 7)

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

25 x^{2} + 9 y^{2} - 225 = 0

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{100} = 1

6 x^{2} + 6 y^{2} + 24 x + 36 y = 0

What is the vertices ((x+2)^2)/(16)+((y+2)^2)/(25)=1 ?
The vertices ((x+2)^2)/(16)+((y+2)^2)/(25)=1 is (-2,3),(-2,-7)