What is the vertices ((x-3)^2)/(25)-((y+4)^2)/(25)=1 ?

The vertices ((x-3)^2)/(25)-((y+4)^2)/(25)=1 is (8,-4),(-2,-4)

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vertices ((x-3)^2)/(25)-((y+4)^2)/(25)=1

vertices

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{25} = 1

(8, - 4), (- 2, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 4), a = 5, b = 5

(3 + 5, - 4), (3 - 5, - 4)

Refine

(8, - 4), (- 2, - 4)

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1

x^{2} - 4 y^{2} - 2 x + 16 y = 20

\frac{( y - 1 ) ^{2}}{16} - \frac{( x + 6 ) ^{2}}{81} = 1

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

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

What is the vertices ((x-3)^2)/(25)-((y+4)^2)/(25)=1 ?
The vertices ((x-3)^2)/(25)-((y+4)^2)/(25)=1 is (8,-4),(-2,-4)