Answer
$${\text{ The equation describes a parabola}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& x = 16{y^2} \cr
& {y^2} = \frac{1}{{16}}x \cr
& \left( a \right) \cr
& {\text{The equation is written in the form }}{y^2} = 4py, \cr
& {\text{then, the equation describes a parabola}}{\text{.}} \cr
& \underbrace {{y^2} = \frac{1}{{16}}x}_{{y^2} = 4py} \to 4p = \frac{1}{{16}},{\text{ }}p = \frac{1}{{64}} \cr
& \cr
& \left( b \right) \cr
& {\text{Characteristics:}} \cr
& {\text{Focus: }}\left( {p,0} \right) \cr
& {\text{Focus: }}\left( {\frac{1}{{64}},0} \right) \cr
& {\text{Vertex: }}\left( {0,0} \right){\text{ }} \cr
& {\text{Directrix: }}x = - p \cr
& {\text{Directrix: }}x = - \frac{1}{{64}} \cr
& \cr
& \left( c \right) \cr
& {\text{The eccentricity of a parabola is 1}} \cr
& e = 1 \cr
& \cr
& \left( d \right) \cr
& {\text{Graph}} \cr} $$
