What is x^2-4y^2+2x+8y-7=0 ?

The solution to x^2-4y^2+2x+8y-7=0 is Hyperbola with (h,k)=(-1,1),a=2,b=1

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x^2-4y^2+2x+8y-7=0

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

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

Solution steps

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

Rewrite

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

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

- 9 x^{2} + 25 y^{2} - 100 y - 125 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{11} = 1

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

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

What is x^2-4y^2+2x+8y-7=0 ?
The solution to x^2-4y^2+2x+8y-7=0 is Hyperbola with (h,k)=(-1,1),a=2,b=1