What is ((x-3)^2)/6-((y-1)^2)/(12)=1 ?

The solution to ((x-3)^2)/6-((y-1)^2)/(12)=1 is Hyperbola with (h,k)=(3,1),a=sqrt(6),b=2sqrt(3)

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((x-3)^2)/6-((y-1)^2)/(12)=1

\frac{( x - 3 ) ^{2}}{6} - \frac{( y - 1 ) ^{2}}{12} = 1

(h, k) = (3, 1), a = 6, b = 23

Solution steps

\frac{( x - 3 ) ^{2}}{6} - \frac{( y - 1 ) ^{2}}{12} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{6} - \frac{( y - 1 ) ^{2}}{12} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, 1), a = 6, b = 23

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

- x^{2} + 4 y^{2} - 16 y + 12 = 0

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{40} = 1

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{13} = 1

10 0^{2} + 15 t^{2} = d^{2}

What is ((x-3)^2)/6-((y-1)^2)/(12)=1 ?
The solution to ((x-3)^2)/6-((y-1)^2)/(12)=1 is Hyperbola with (h,k)=(3,1),a=sqrt(6),b=2sqrt(3)