What is ((x-3)^2)/4-((y-2)^2)/(12)=1 ?

The solution to ((x-3)^2)/4-((y-2)^2)/(12)=1 is Hyperbola with (h,k)=(3,2),a=2,b=2sqrt(3)

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((x-3)^2)/4-((y-2)^2)/(12)=1

\frac{( x - 3 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{12} = 1

(h, k) = (3, 2), a = 2, b = 23

Solution steps

\frac{( x - 3 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{12} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{12} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, 2), a = 2, b = 23

4 x^{2} - 9 y^{2} - 16 x + 18 y + 11 = 0

\frac{( y + 1 ) ^{2}}{9} - \frac{( x + 3 ) ^{2}}{4} = 1

16 x^{2} - 64 x - 9 y^{2} - 54 y + 1 = 0

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

7 x^{2} - 9 y^{2} - 63 = 0

What is ((x-3)^2)/4-((y-2)^2)/(12)=1 ?
The solution to ((x-3)^2)/4-((y-2)^2)/(12)=1 is Hyperbola with (h,k)=(3,2),a=2,b=2sqrt(3)