What is the vertices ((x+1)^2)/9+((y-3)^2)/(16)=1 ?

The vertices ((x+1)^2)/9+((y-3)^2)/(16)=1 is (-1,7),(-1,-1)

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vertices ((x+1)^2)/9+((y-3)^2)/(16)=1

vertices

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 1, 3), b = 4, a = 3

(- 1, 3 + 4), (- 1, 3 - 4)

Refine

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

16 x^{2} - 96 x + 9 y^{2} = 0

4 x^{2} + 25 y^{2} - 50 y = 75

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

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

4 x^{2} + 9 y^{2} - 36 = 0

What is the vertices ((x+1)^2)/9+((y-3)^2)/(16)=1 ?
The vertices ((x+1)^2)/9+((y-3)^2)/(16)=1 is (-1,7),(-1,-1)