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

The vertices ((x-2)^2)/5+((y+3)^2)/9 =1 is (2,0),(2,-6)

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

vertices

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

(2, 0), (2, - 6)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(2, 0), (2, - 6)

25 x^{2} + y^{2} - 4 y - 21 = 0

9 x^{2} - 72 x + y^{2} + 4 y + 112 = 0

4 x^{2} + 9 y^{2} = 1

4 x^{2} + y^{2} = 4

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0

What is the vertices ((x-2)^2)/5+((y+3)^2)/9 =1 ?
The vertices ((x-2)^2)/5+((y+3)^2)/9 =1 is (2,0),(2,-6)