What is the vertices ((y+1)^2)/(25)-((x+4)^2)/(64)=1 ?

The vertices ((y+1)^2)/(25)-((x+4)^2)/(64)=1 is (-4,4),(-4,-6)

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vertices ((y+1)^2)/(25)-((x+4)^2)/(64)=1

vertices

\frac{( y + 1 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{64} = 1

(- 4, 4), (- 4, - 6)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 1 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{64} = 1 :

Up-down Hyperbola with

(h, k) = (- 4, - 1), a = 5, b = 8

(- 4, - 1 + 5), (- 4, - 1 - 5)

Refine

(- 4, 4), (- 4, - 6)

x^{2} - y^{2} - 6 x = 16

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

\frac{x ^{2}}{45} - \frac{y ^{2}}{360} = 1

49 x^{2} - 25 y^{2} - 686 x - 50 y + 2375 = 0

What is the vertices ((y+1)^2)/(25)-((x+4)^2)/(64)=1 ?
The vertices ((y+1)^2)/(25)-((x+4)^2)/(64)=1 is (-4,4),(-4,-6)