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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

2 x^{2} - 2 y^{2} + 3 x - 2 y - 43 = 0

\frac{30}{x + y} + \frac{54}{x - y} = 8

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

y^{2} - 5 x^{2} = 20

\frac{( x ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{16} = 1

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