What is the vertices ((x-3)^2)/4+((y+2)^2)/1 =1 ?

The vertices ((x-3)^2)/4+((y+2)^2)/1 =1 is (5,-2),(1,-2)

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vertices ((x-3)^2)/4+((y+2)^2)/1 =1

vertices

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

(5, - 2), (1, - 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (3, - 2), a = 2, b = 1

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

Refine

(5, - 2), (1, - 2)

\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

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

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

What is the vertices ((x-3)^2)/4+((y+2)^2)/1 =1 ?
The vertices ((x-3)^2)/4+((y+2)^2)/1 =1 is (5,-2),(1,-2)