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

The vertices ((x-5)^2)/4-((y+5)^2)/9 =1 is (7,-5),(3,-5)

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

vertices

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

(7, - 5), (3, - 5)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(5 + 2, - 5), (5 - 2, - 5)

Refine

(7, - 5), (3, - 5)

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

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

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

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

\frac{y ^{2}}{25} - \frac{x ^{2}}{144} = 1

What is the vertices ((x-5)^2)/4-((y+5)^2)/9 =1 ?
The vertices ((x-5)^2)/4-((y+5)^2)/9 =1 is (7,-5),(3,-5)