Explanation

(c)
The given equation of parabola is
$y = {2x}^{2} + x\ \Rightarrow \ x^{2} + \frac{x}{2} = \frac{y}{2}$
$\Rightarrow x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{y}{2} + \frac{1}{16}$
$\Rightarrow \left( x + \frac{1}{4} \right)^{2} = \frac{1}{2}\left( y + \frac{1}{8} \right)$
It can be rewritten as $X^{2} = \frac{1}{2}Y$ …(i)
Where $x + \frac{1}{4} = X$ and $y + \frac{1}{8} = Y$
On comparing with $X^{2} = 4AY$, we get
$A = \frac{1}{8}$, focus of Eq. (i) is $\left( 0,\ \frac{1}{8} \right)$ $ie$,
$X = 0,\ Y = \frac{1}{8}$
$\Rightarrow \ x + \frac{1}{4} = 0,\ y + \frac{1}{8} = \frac{1}{8}$
$\Rightarrow x = - \frac{1}{4},\ y = 0$
$\therefore$ Focus of given parabola is $\left( - \frac{1}{4},\ 0 \right)$