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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

a^{2} + b^{2} i^{2} = 9

8 x^{2} - 2 y^{2} - 16 x - 16 y - 32 = 0

9 x^{2} - 25 y^{2} - 90 x + 50 y - 25 = 0

\frac{( y + 5 ) ^{2}}{81} - \frac{( x + 8 ) ^{2}}{9} = 1

x^{2} - y^{2} - 1 = 3 x - 3 y

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