What is -x^2+y^2-4x-13=0 ?

The solution to -x^2+y^2-4x-13=0 is Hyperbola with (h,k)=(-2,0),a=3,b=3

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-x^2+y^2-4x-13=0

- x^{2} + y^{2} - 4 x - 13 = 0

(h, k) = (- 2, 0), a = 3, b = 3

Solution steps

- x^{2} + y^{2} - 4 x - 13 = 0

Rewrite

- x^{2} + y^{2} - 4 x - 13 = 0

in the form of a standard hyperbola equation

\frac{( y - 0 ) ^{2}}{3 ^{2}} - \frac{( x - ( - 2 ) ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, 0), a = 3, b = 3

z^{2} + 1 0^{2} = 2 + q^{2} + \frac{9 6 ^{2}}{2}

(x^{2} + 2 x - 2 y^{2}) = 0

\frac{( x - 4 ) ^{2}}{25} - \frac{( y + 8 ) ^{2}}{49} = 1

\frac{( x - 4 ) ^{2}}{9} - \frac{( y + 5 ) ^{2}}{16} = 1

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

What is -x^2+y^2-4x-13=0 ?
The solution to -x^2+y^2-4x-13=0 is Hyperbola with (h,k)=(-2,0),a=3,b=3