What is the center ((y-1)^2)/9-((x+3)^2)/(16)=1 ?

The center ((y-1)^2)/9-((x+3)^2)/(16)=1 is (-3,1)

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center ((y-1)^2)/9-((x+3)^2)/(16)=1

center

\frac{( y - 1 ) ^{2}}{9} - \frac{( x + 3 ) ^{2}}{16} = 1

(- 3, 1)

Solution steps

\frac{( y - 1 ) ^{2}}{9} - \frac{( x + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( y - 1 ) ^{2}}{9} - \frac{( x + 3 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 3, 1), a = 3, b = 4

And the center is:

(- 3, 1)

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

\frac{1}{144} x^{2} - \frac{1}{196} y^{2} = 1

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0^{2}

\frac{( x + 1 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{25} = 1

10 x^{2} - 25 y^{2} = 100

What is the center ((y-1)^2)/9-((x+3)^2)/(16)=1 ?
The center ((y-1)^2)/9-((x+3)^2)/(16)=1 is (-3,1)