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

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

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

vertices

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

(- 1, 7), (- 1, - 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 2), b = 5, a = 4

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

Refine

(- 1, 7), (- 1, - 3)

x^{2} + 2 y^{2} - 1 = 0

9 x^{2} + 25 y^{2} = 9

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

81 x^{2} + 4 y^{2} = 324

x^{2} - 4 x + 100 y^{2} + 200 y + 4 = 0

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