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

The vertices ((x-2)^2)/4-((y+3)^2)/5 =1 is (4,-3),(0,-3)

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

vertices

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{5} = 1

(4, - 3), (0, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{5} = 1 :

Right-left Hyperbola with

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

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

Refine

(4, - 3), (0, - 3)

\frac{( x - 8 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

25 y^{2} - x^{2} = 100

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

9 x^{2} - y^{2} - 81 = 0

16 x^{2} - 12 y^{2} = 192

What is the vertices ((x-2)^2)/4-((y+3)^2)/5 =1 ?
The vertices ((x-2)^2)/4-((y+3)^2)/5 =1 is (4,-3),(0,-3)