What is the vertices ((x-2)^2)/(36)+((y+1)^2)/(25)=1 ?

The vertices ((x-2)^2)/(36)+((y+1)^2)/(25)=1 is (8,-1),(-4,-1)

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vertices ((x-2)^2)/(36)+((y+1)^2)/(25)=1

vertices

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

(8, - 1), (- 4, - 1)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (2, - 1), a = 6, b = 5

(2 + 6, - 1), (2 - 6, - 1)

Refine

(8, - 1), (- 4, - 1)

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{25} = 1

2 \frac{( x - 6 ) ^{2}}{3} + 7 \frac{( y + 4 ) ^{2}}{4} = 1

\frac{x ^{2}}{96} + \frac{y ^{2}}{100} = 1

\frac{( x + 4 ) ^{2}}{24} + \frac{( y + 2 ) ^{2}}{49} = 1

What is the vertices ((x-2)^2)/(36)+((y+1)^2)/(25)=1 ?
The vertices ((x-2)^2)/(36)+((y+1)^2)/(25)=1 is (8,-1),(-4,-1)