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

The vertices ((x+2)^2)/4-((y-1)^2)/9 =1 is (0,1),(-4,1)

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

vertices

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

(0, 1), (- 4, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

(0, 1), (- 4, 1)

(4 x - 3 y) = \frac{- 144}{( 4 x + 3 y )}

4 y^{2} - 5 x^{2} = 20

\frac{y ^{2}}{25} - \frac{x ^{2}}{49} = 1

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

- x^{2} + 4 y^{2} + 2 x + 24 y + 31 = 0

What is the vertices ((x+2)^2)/4-((y-1)^2)/9 =1 ?
The vertices ((x+2)^2)/4-((y-1)^2)/9 =1 is (0,1),(-4,1)