What is the vertices ((x-8)^2)/9-((y+3)^2)/(16)=1 ?

The vertices ((x-8)^2)/9-((y+3)^2)/(16)=1 is (11,-3),(5,-3)

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vertices ((x-8)^2)/9-((y+3)^2)/(16)=1

vertices

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

(11, - 3), (5, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (8, - 3), a = 3, b = 4

(8 + 3, - 3), (8 - 3, - 3)

Refine

(11, - 3), (5, - 3)

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

x^{2} - y^{2} = 14

What is the vertices ((x-8)^2)/9-((y+3)^2)/(16)=1 ?
The vertices ((x-8)^2)/9-((y+3)^2)/(16)=1 is (11,-3),(5,-3)