What is (x^2-4x)-(y^2+4y)=1 ?

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

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(x^2-4x)-(y^2+4y)=1

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

(h, k) = (2, - 2), a = 1, b = 1

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 2), a = 1, b = 1

6 x^{2} - 3 y^{2} + 36 x + 30 y = 39

y^{2} - 4 = 9 x^{2}

(4 - 4 x)^{2} - 4 (2 k (x) + 2) (k (x) - 2) = 0

4 x^{2} - y^{2} - 32 x - 2 y = - 59

x^{2} - y^{2} = - 45

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