What is the vertices ((x+1)^2)/9+((y-3)^2)/(10)=1 ?

The vertices ((x+1)^2)/9+((y-3)^2)/(10)=1 is (-1,3+sqrt(10)),(-1,3-sqrt(10))

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vertices ((x+1)^2)/9+((y-3)^2)/(10)=1

vertices

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1

(- 1, 3 + 10), (- 1, 3 - 10)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1 :

Ellipse with center

(h, k) = (- 1, 3), b = 10, a = 3

(- 1, 3 + 10), (- 1, 3 - 10)

\frac{x ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

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

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

\frac{x ^{2}}{5} + \frac{y ^{2}}{9} = 1

\frac{( x + 4 ) ^{2}}{5} + \frac{( y - 6 ) ^{2}}{9} = 1

What is the vertices ((x+1)^2)/9+((y-3)^2)/(10)=1 ?
The vertices ((x+1)^2)/9+((y-3)^2)/(10)=1 is (-1,3+sqrt(10)),(-1,3-sqrt(10))