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

The vertices ((x-1)^2)/4+((y+2)^2)/9 =1 is (1,1),(1,-5)

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(1, - 2 + 3), (1, - 2 - 3)

Refine

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

16 x^{2} - 32 x + 9 y^{2} = 128

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

5 x^{2} + 2 y^{2} = 10

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

4 x^{2} + 16 y^{2} - 64 = 0

What is the vertices ((x-1)^2)/4+((y+2)^2)/9 =1 ?
The vertices ((x-1)^2)/4+((y+2)^2)/9 =1 is (1,1),(1,-5)