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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

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

4 x^{2} - 6 y^{2} = 24

9 x^{2} - 2 y^{2} = 18

7^{2} + b^{2} = (c + 1)^{2}

- 16 x^{2} + 9 y^{2} - 144 = 0

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