What is x^2+2x-y^2+1-4=0 ?

The solution to x^2+2x-y^2+1-4=0 is Hyperbola with (h,k)=(-1,0),a=2,b=2

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x^2+2x-y^2+1-4=0

x^{2} + 2 x - y^{2} + 1 - 4 = 0

(h, k) = (- 1, 0), a = 2, b = 2

Solution steps

x^{2} + 2 x - y^{2} + 1 - 4 = 0

Rewrite

x^{2} + 2 x - y^{2} + 1 - 4 = 0

in the form of a standard hyperbola equation

\frac{( x - ( - 1 ) ) ^{2}}{2 ^{2}} - \frac{( y - 0 ) ^{2}}{2 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 1, 0), a = 2, b = 2

8 x^{2} - 4 y^{2} + 10 x - 6 y - 8 = 0

\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1

2 x^{2} - y^{2} - 1 = 0

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{5} = 1

What is x^2+2x-y^2+1-4=0 ?
The solution to x^2+2x-y^2+1-4=0 is Hyperbola with (h,k)=(-1,0),a=2,b=2