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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

9 x^{2} - 100 y^{2} = 25

4 x^{2} - 49 y^{2} - 24 x + 882 y - 4129 = 0

\frac{- y ^{2}}{4} + \frac{z ^{2}}{4} = 1 - \frac{5}{4}

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

m, x (x + 17) < 60 m (x)^{2}

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