Standard Equation of Hyperbola

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Write the equation of the *hyperbola* in standard from $$25x^{2}-4y^{2}+150x-48y+181 = 0 $$

**Answer**

$25x^{2}-4y^{2}+150x-48y+181 = 0 $

$25x^{2}+150x-4y^{2}-48y=-181$

$25(x^{2}+6x)-4(y^{2}+12y)=-181$

$25(x^{2}+6x+9)-4(y^{2}+12y+36)=-181+225-144$

$25(x+3)^2-4(y+6)^2 = -100$

$-\frac{1}{4}(x+3)^2+\frac{1}{25}(y+6)^2 = 1$

$\frac{(y+6)^2}{{5}^2}-\frac{(x+3)^2}{{2}^2} =1$