What is -x^2+4x+3y^2-12y-4=0 ?

The solution to -x^2+4x+3y^2-12y-4=0 is Hyperbola with (h,k)=(2,2),a=2,b=2sqrt(3)

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-x^2+4x+3y^2-12y-4=0

- x^{2} + 4 x + 3 y^{2} - 12 y - 4 = 0

(h, k) = (2, 2), a = 2, b = 23

Solution steps

- x^{2} + 4 x + 3 y^{2} - 12 y - 4 = 0

Rewrite

- x^{2} + 4 x + 3 y^{2} - 12 y - 4 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, 2), a = 2, b = 23

16 \cdot (x^{2} - 2 x) - 25 (y^{2} - 4 y) = 484

\frac{( x - 2 ) ^{2}}{12} - \frac{( 3 y ^{2} + y )}{12} = 1

2 x - 4 x^{2} = - 9 y^{2} - 21 y - 50

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{4} = 1

(\frac{K ( x )}{2})^{2} = 4096 + 4096 - 128 x + x^{2}

What is -x^2+4x+3y^2-12y-4=0 ?
The solution to -x^2+4x+3y^2-12y-4=0 is Hyperbola with (h,k)=(2,2),a=2,b=2sqrt(3)