What is ((x-3)^2)/(36)-((y+2)^2)/(64)=1 ?

The solution to ((x-3)^2)/(36)-((y+2)^2)/(64)=1 is Hyperbola with (h,k)=(3,-2),a=6,b=8

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((x-3)^2)/(36)-((y+2)^2)/(64)=1

\frac{( x - 3 ) ^{2}}{36} - \frac{( y + 2 ) ^{2}}{64} = 1

(h, k) = (3, - 2), a = 6, b = 8

Solution steps

\frac{( x - 3 ) ^{2}}{36} - \frac{( y + 2 ) ^{2}}{64} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{36} - \frac{( y + 2 ) ^{2}}{64} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{6 ^{2}} - \frac{( y - ( - 2 ) ) ^{2}}{8 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, - 2), a = 6, b = 8

3 x^{2} - 4 y^{2} = 7

(4 x - 2 y) (6 x + 3 y) = 18

9 x^{2} - 4 y^{2} - 36 x - 8 y - 4 = 0

36 x^{2} - 9 y^{2} = 324

\frac{1}{4} (x - 2)^{2} - \frac{1}{16} (y + 7)^{2} = 1

What is ((x-3)^2)/(36)-((y+2)^2)/(64)=1 ?
The solution to ((x-3)^2)/(36)-((y+2)^2)/(64)=1 is Hyperbola with (h,k)=(3,-2),a=6,b=8