What is the vertices ((x-1)^2)/(25)-((y+6)^2)/(49)=1 ?

The vertices ((x-1)^2)/(25)-((y+6)^2)/(49)=1 is (6,-6),(-4,-6)

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vertices ((x-1)^2)/(25)-((y+6)^2)/(49)=1

vertices

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 6 ) ^{2}}{49} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 6 ) ^{2}}{49} = 1 :

Right-left Hyperbola with

(h, k) = (1, - 6), a = 5, b = 7

(1 + 5, - 6), (1 - 5, - 6)

Refine

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

4 x^{2} - 24 x - 36 y^{2} - 360 y + 864 = 0

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

\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

What is the vertices ((x-1)^2)/(25)-((y+6)^2)/(49)=1 ?
The vertices ((x-1)^2)/(25)-((y+6)^2)/(49)=1 is (6,-6),(-4,-6)