What is x^2-4y^2+2x-32y-67=0 ?

The solution to x^2-4y^2+2x-32y-67=0 is Hyperbola with (h,k)=(-1,-4),a=2,b=1

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x^2-4y^2+2x-32y-67=0

x^{2} - 4 y^{2} + 2 x - 32 y - 67 = 0

(h, k) = (- 1, - 4), a = 2, b = 1

Solution steps

x^{2} - 4 y^{2} + 2 x - 32 y - 67 = 0

Rewrite

x^{2} - 4 y^{2} + 2 x - 32 y - 67 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 1, - 4), a = 2, b = 1

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

\frac{( y - 4 ) ^{2}}{100} - \frac{( x - 7 ) ^{2}}{49} = 1

- x^{2} + 9 y^{2} - 54 y + 45 = 0

9 x^{2} - 25 y^{2} - 90 x - 100 y - 100 = 0

\frac{( y ^{2} )}{25} - \frac{( x ^{2} )}{9} = 1

What is x^2-4y^2+2x-32y-67=0 ?
The solution to x^2-4y^2+2x-32y-67=0 is Hyperbola with (h,k)=(-1,-4),a=2,b=1