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

The solution to ((x-3)^2}{16}-\frac{(y+1)^2)/9 =1 is Hyperbola with (h,k)=(3,-1),a=4,b=3

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

(\frac{y}{100})^{2} - \frac{x ^{2}}{49} = 1

(x + y) (x - y) = 60.49

9 x^{2} - 4 y^{2} = 144

9 x^{2} - 16 y^{2} - 54 x + 32 y - 79 = 0

\frac{( x - 6 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{9} = 1

What is ((x-3)^2}{16}-\frac{(y+1)^2)/9 =1 ?
The solution to ((x-3)^2}{16}-\frac{(y+1)^2)/9 =1 is Hyperbola with (h,k)=(3,-1),a=4,b=3