<s>[INST] What is the eccentricity of $4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$? Question: What is the eccentricity of $4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$?
Source: James S. Rickards Invitational (Algebra II Individual)
I know how to calculate eccentricity, but I am having trouble on factoring it in to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ form. 
This is what I did: 
$4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$
$4\sqrt{5}x^2 − 17x + y=-\frac{1}{8}$
$4\sqrt{5}(x^2-\frac{17x}{4\sqrt{5}})+1(y+0)= -\frac{1}{8}$
$(x-\frac{17\sqrt{5}}{40})^2+\frac{y}{4\sqrt5}=\frac{-32\sqrt5}{5120}-\frac{289}{160}$
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 [/INST] Note, as given, you have: $4\sqrt{5}x^2-17x+y+\frac{1}{8}=0$.
Just rewrite this as $y=-4\sqrt{5}x^2+17x-\frac{1}{8}$.
It is very easy to see now that this is just the equation of a parabola. 
Now by definition, any parabola has an eccentricity of $1$. 
No need to complete the square here. 
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