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

The solution to x^2-y^2+4x+2y-4=0 is Hyperbola with (h,k)=(-2,1),a=sqrt(7),b=sqrt(7)

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

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

(h, k) = (- 2, 1), a = 7, b = 7

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - ( - 2 ) ) ^{2}}{( 7 ) ^{2}} - \frac{( y - 1 ) ^{2}}{( 7 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, 1), a = 7, b = 7

x^{2} - 8 y^{2} - 6 x - 22 y + 4 = 0

\frac{( y - 11 ) ^{2}}{16} - \frac{( x + 5 ) ^{2}}{36} = 1

- 25 x^{2} + y^{2} = 25

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

a = \frac{2020 - c}{\frac{2021 - a}{c}}

What is x^2-y^2+4x+2y-4=0 ?
The solution to x^2-y^2+4x+2y-4=0 is Hyperbola with (h,k)=(-2,1),a=sqrt(7),b=sqrt(7)