Respuesta :

check the picture below.

[tex]\bf \textit{ellipse, vertical major axis}\\\\ \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ ------\\ h=0\\ k=0 \end{cases} \\\\\\ \cfrac{(x-0)^2}{2^2}+\cfrac{(y-0)^2}{4^2}=1\implies \cfrac{x^2}{4}+\cfrac{y^2}{16}=1[/tex]
Ver imagen jdoe0001