拋物線 $y^2 = 2x$ 之 $4p = 2 \implies p = \dfrac{1}{2}$，頂點 $(0,0)$，故焦點為 $\left( \dfrac{1}{2}, 0 \right)$。 $(1)$ $y + \dfrac{1}{4} = \left( x - \dfrac{1}{2} \right)^2$ 為開口向上之拋物線，$4p = 1 \implies p = \dfrac{1}{4}$，頂點 $\left( \dfrac{1}{2}, -\dfrac{1}{4} \right)$，焦點為 $\left( \dfrac{1}{2}, -\dfrac{1}{4} + \dfrac{1}{4} \right) = \left( \dfrac{1}{2}, 0 \right)$。（正確） $(2)$ $\dfrac{x^2}{4} + \dfrac{y^2}{3} = 1$ 為橢圓，$a^2=4, b^2=3 \implies c^2=1 \implies c=1$，焦點 $(\pm 1, 0)$。 $(3)$ $\dfrac{x^2}{1} + \dfrac{y^2}{3/4} = 1$ 為橢圓，$a^2=1, b^2=\dfrac{3}{4} \implies c^2=\dfrac{1}{4} \implies c=\dfrac{1}{2}$，焦點為 $\left( \pm \dfrac{1}{2}, 0 \right)$。（正確） $(4)$ $\dfrac{x^2}{1/8} - \dfrac{y^2}{1/8} = 1$ 為雙曲線，$a^2=\dfrac{1}{8}, b^2=\dfrac{1}{8} \implies c^2=a^2+b^2=\dfrac{1}{4} \implies c=\dfrac{1}{2}$，焦點為 $\left( \pm \dfrac{1}{2}, 0 \right)$。（正確） $(5)$ $\dfrac{x^2}{1/4} - \dfrac{y^2}{1/4} = 1$ 為雙曲線，$a^2=\dfrac{1}{4}, b^2=\dfrac{1}{4} \implies c^2=\dfrac{1}{2} \implies c=\dfrac{1}{\sqrt{2}}$。 故選 $(1)(3)(4)$。