Standard Equation of Hyperbola

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Write the equation of the *hyperbola* in standard from $$x^{2}-4y^{2}+2x-32y-67 = 0 $$

**Answer**

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

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

$(x^{2}+2x)-4(y^{2}+8y)=67$

$(x^{2}+2x+1)-4(y^{2}+8y+16)=67+1-64$

$(x+1)^2-4(y+4)^2 = 4$

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

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