Standard Equation of Hyperbola

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Write the equation of the *hyperbola* in standard from $$x^{2}-y^{2}+12x+8y+45 = 0 $$

**Answer**

$x^{2}-y^{2}+12x+8y+45 = 0 $

$x^{2}+12x-y^{2}+8y=-45$

$(x^{2}+12x)-(y^{2}-8y)=-45$

$(x^{2}+12x+36)-(y^{2}-8y+16)=-45+36-16$

$(x+6)^2-(y-4)^2 = -25$

$-\frac{1}{25}(x+6)^2+\frac{1}{25}(y-4)^2 = 1$

$\frac{(y-4)^2}{{5}^2}-\frac{(x+6)^2}{{5}^2} =1$