What is the vertices ((x-5)^2)/4-((y+1)^2)/(49)=1 ?

The vertices ((x-5)^2)/4-((y+1)^2)/(49)=1 is (7,-1),(3,-1)

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vertices ((x-5)^2)/4-((y+1)^2)/(49)=1

vertices

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{49} = 1

(7, - 1), (3, - 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{49} = 1 :

Right-left Hyperbola with

(h, k) = (5, - 1), a = 2, b = 7

(5 + 2, - 1), (5 - 2, - 1)

Refine

(7, - 1), (3, - 1)

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

\frac{y ^{2}}{25} - \frac{x ^{2}}{81} = 1

9 x^{2} - 18 x - 16 y^{2} - 64 y = 199

\frac{y ^{2}}{49} - \frac{x ^{2}}{16} = 1

What is the vertices ((x-5)^2)/4-((y+1)^2)/(49)=1 ?
The vertices ((x-5)^2)/4-((y+1)^2)/(49)=1 is (7,-1),(3,-1)