What is the vertices ((x+4)^2)/9-((y-8)^2)/(49)=1 ?

The vertices ((x+4)^2)/9-((y-8)^2)/(49)=1 is (-1,8),(-7,8)

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vertices ((x+4)^2)/9-((y-8)^2)/(49)=1

vertices

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 8 ) ^{2}}{49} = 1

(- 1, 8), (- 7, 8)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 8 ) ^{2}}{49} = 1 :

Right-left Hyperbola with

(h, k) = (- 4, 8), a = 3, b = 7

(- 4 + 3, 8), (- 4 - 3, 8)

Refine

(- 1, 8), (- 7, 8)

25 x^{2} - 9 y^{2} - 225 = 0

5 x^{2} - 4 y^{2} - 40 x + 24 y + 24 = 0

x^{2} - y^{2} - 4 x - 6 y - 14 = 0

x^{2} - 16 y^{2} = 64

\frac{x ^{2}}{4} - \frac{y ^{2}}{81} = 1

What is the vertices ((x+4)^2)/9-((y-8)^2)/(49)=1 ?
The vertices ((x+4)^2)/9-((y-8)^2)/(49)=1 is (-1,8),(-7,8)