What is the vertices ((x-3)^2)/(49)-((y+5)^2)/(25)=1 ?

The vertices ((x-3)^2)/(49)-((y+5)^2)/(25)=1 is (10,-5),(-4,-5)

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vertices ((x-3)^2)/(49)-((y+5)^2)/(25)=1

vertices

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1

(10, - 5), (- 4, - 5)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 5), a = 7, b = 5

(3 + 7, - 5), (3 - 7, - 5)

Refine

(10, - 5), (- 4, - 5)

9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

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

\frac{x ^{2}}{4} - \frac{y ^{2}}{25} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

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

What is the vertices ((x-3)^2)/(49)-((y+5)^2)/(25)=1 ?
The vertices ((x-3)^2)/(49)-((y+5)^2)/(25)=1 is (10,-5),(-4,-5)