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

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

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

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

(h, k) = (2, - 1), a = 4, b = 3

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 1), a = 4, b = 3

9 x^{2} - 16 y^{2} - 108 x + 64 y + 116 = 0

\frac{( y + 3 ) ^{2}}{2} - \frac{( x - 1 ) ^{2}}{18} = 1

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

x^{2} - 3 y^{2} + 11 x + y + 10 = 0

16 y^{2} - 4 x^{2} - 24 x - 12 y - 34 = 0

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