What is (3x-3)^2-(2y-4)^2=68 ?

The solution to (3x-3)^2-(2y-4)^2=68 is Hyperbola with (h,k)=(1,2),a=(2sqrt(17))/3 ,b=sqrt(17)

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(3x-3)^2-(2y-4)^2=68

(3 x - 3)^{2} - (2 y - 4)^{2} = 68

(h, k) = (1, 2), a = \frac{2 17}{3}, b = 17

Solution steps

(3 x - 3)^{2} - (2 y - 4)^{2} = 68

Rewrite

(3 x - 3)^{2} - (2 y - 4)^{2} = 68

in the form of a standard hyperbola equation

\frac{( x - 1 ) ^{2}}{( \frac{2 17}{3} ) ^{2}} - \frac{( y - 2 ) ^{2}}{( 17 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (1, 2), a = \frac{2 17}{3}, b = 17

64 x^{2} - 100 y^{2} = 6400

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

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

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

What is (3x-3)^2-(2y-4)^2=68 ?
The solution to (3x-3)^2-(2y-4)^2=68 is Hyperbola with (h,k)=(1,2),a=(2sqrt(17))/3 ,b=sqrt(17)