Chapter 10 - Parametric and Polar Curves - 10.4 Conic Sections - 10.4 Exercises - Page 750: 22

$${y^2} = - 16x$$

$$\eqalign{ & {\text{The focus at }}\left( { - 4} \right){\text{ and Vertex }}\left( {0,0} \right) \cr & {\text{Then the parabola has the equation }}{y^2} = 4px \cr & {\text{With focus }}\left( {p,0} \right) \cr & \left( { - 4,0} \right)\,\, \Rightarrow \,\,\,\,\,p = - 4 \cr & {\text{The equation of the parabola is:}} \cr & {y^2} = 4\left( { - 4} \right)x \cr & {y^2} = - 16x \cr} $$