Respuesta :

[tex]\bf \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -----------------------------\\\\ so\quad \begin{cases} \textit{the equation went from}\\\\ (y-4)^2=-4(x+3)\\\\ to\\\\ (y-4)^2=-4(x+10) \end{cases}[/tex]

notice the h,k vertex of the first one
notice the second one, notice how "h" changed