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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

9 x^{2} - 4 y^{2} + 54 x + 16 y + 29 = 0

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

\frac{( y - 0 ) ^{2}}{3 ^{2}} - \frac{( x - 2 ) ^{2}}{4 ^{2}} = 1

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

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

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