What is the vertices ((x-3)^2)/4+((y+1)^2)/9 =1 ?

The vertices ((x-3)^2)/4+((y+1)^2)/9 =1 is (3,2),(3,-4)

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vertices ((x-3)^2)/4+((y+1)^2)/9 =1

vertices

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

(3, 2), (3, - 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(3, 2), (3, - 4)

16 x^{2} - 96 x + 144 + 25 y^{2} = 400

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

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

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

25 x^{2} + 49 y^{2} = 1225

What is the vertices ((x-3)^2)/4+((y+1)^2)/9 =1 ?
The vertices ((x-3)^2)/4+((y+1)^2)/9 =1 is (3,2),(3,-4)