What is ((x-3)^2)/4-((y+1)^2)/(36)=1 ?

The solution to ((x-3)^2)/4-((y+1)^2)/(36)=1 is Hyperbola with (h,k)=(3,-1),a=2,b=6

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((x-3)^2)/4-((y+1)^2)/(36)=1

\frac{( x - 3 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{36} = 1

(h, k) = (3, - 1), a = 2, b = 6

Solution steps

\frac{( x - 3 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{36} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{2 ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{6 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, - 1), a = 2, b = 6

x^{2} - y^{2} + 4 y = 0

\frac{x ^{2}}{65} - \frac{y ^{2}}{65} = 1

x^{2} - y^{2} + 4 y = 1

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

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

What is ((x-3)^2)/4-((y+1)^2)/(36)=1 ?
The solution to ((x-3)^2)/4-((y+1)^2)/(36)=1 is Hyperbola with (h,k)=(3,-1),a=2,b=6