The focus of the parabola \(y=2 x^{2}+x\) is

The given equation of parabola is
\(y=2 x^{2}+x\) \(\Rightarrow\) \(x^{2}+\frac{x}{2}=\frac{y}{2}\) \(\Rightarrow \quad x^{2}+\frac{x}{2}+\frac{1}{16}=\frac{y}{2}+\frac{1}{16}\) \(\Rightarrow \quad\left(x+\frac{1}{4}\right)^{2}=\frac{1}{2}\left(y+\frac{1}{8}\right)\)
where \(x+\frac{1}{4}=X\) and \(y+\frac{1}{8}=Y\)
On comparing with \(X^{2}=4 A Y\), we get
\[
A=\frac{1}{8}
\]
focus of Eq. (i) is \(\left(0, \frac{1}{8}\right)\) i.e. \(X=0, Y=\frac{1}{8}\) \(\Rightarrow x+\frac{1}{4}=0, y+\frac{1}{8}=\frac{1}{8} \Rightarrow x=-\frac{1}{4}, y=0\)
\(\therefore\) Focus of given parabola is \(\left(-\frac{1}{4}, 0\right)\).