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

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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

(5, 3), (- 1, 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

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

9 x^{2} + y^{2} = 84

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 = 0

3 x^{2} + 7 y^{2} = 63

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

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