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

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

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

vertices

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

(8, - 1), (- 2, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (3, - 1), a = 5, b = 25

(3 + 5, - 1), (3 - 5, - 1)

Refine

(8, - 1), (- 2, - 1)

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 4 ) ^{2}}{25} = 1

10 y^{2} - 5 x^{2} = 50

\frac{x ^{2}}{9} - \frac{y ^{2}}{36} = 1

4 x^{2} - y^{2} = 4

2 x^{2} - 2 y^{2} - 20 x + 16 y - 16 = 0

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