What is ((x-3)^2}{25}-\frac{(y-1)^2)/4 =1 ?

The solution to ((x-3)^2}{25}-\frac{(y-1)^2)/4 =1 is Hyperbola with (h,k)=(3,1),a=5,b=2

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((x-3)^2}{25}-\frac{(y-1)^2)/4 =1

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{4} = 1

(h, k) = (3, 1), a = 5, b = 2

Solution steps

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{4} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{5 ^{2}} - \frac{( y - 1 ) ^{2}}{2 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, 1), a = 5, b = 2

\frac{x ^{2}}{7} - \frac{y ^{2}}{6} = 1

25 x^{2} - y^{2} + 50 x + 6 y - 9 = 0

\frac{( x + 1 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{9} = 1

x^{2} - y^{2} - 1 < 0

\frac{y ^{2}}{9} - \frac{x ^{2}}{1} = 1

What is ((x-3)^2}{25}-\frac{(y-1)^2)/4 =1 ?
The solution to ((x-3)^2}{25}-\frac{(y-1)^2)/4 =1 is Hyperbola with (h,k)=(3,1),a=5,b=2