What is the vertices ((x+2)^2)/9+((y-3)^2)/5 =1 ?

The vertices ((x+2)^2)/9+((y-3)^2)/5 =1 is (1,3),(-5,3)

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vertices ((x+2)^2)/9+((y-3)^2)/5 =1

vertices

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

(1, 3), (- 5, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (- 2, 3), a = 3, b = 5

(- 2 + 3, 3), (- 2 - 3, 3)

Refine

(1, 3), (- 5, 3)

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

9 x^{2} + y^{2} - 18 x - 6 y + 9 = 0

\frac{( x - 0 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

81 x^{2} + 64 y^{2} = 5184

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

What is the vertices ((x+2)^2)/9+((y-3)^2)/5 =1 ?
The vertices ((x+2)^2)/9+((y-3)^2)/5 =1 is (1,3),(-5,3)