we must simplify it.
![2(x-\frac{1}{4})^2 -\frac{1}{8}+4y^2 =16\\\\2(x-\frac{1}{4})^2+4y^2 =\frac{129}{8}\\\\\frac{8}{129}[2(x-\frac{1}{4})^2] +\frac{8}{129}[4y^2] =1\\\\\frac{16(x-\frac{1}{4})^2}{129}+\frac{32(y-0)^2}{129}=1](/tpl/images/0402/8351/16284.png)
 and  
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Mathematics, 04.12.2019 12:31 jbrown76241
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