Question 665233
{{{(x+2)^2/4+(y+3)^2/16=1}}}


the major axis is {{{vertical}}} (because {{{4}}} is smaller than {{{16}}})

so, the {{{sqrt(16) =4}}}

the {{{center}}} is at ({{{-2}}},{{{-3}}}), so move over {{{4}}} vertically on both sides, and those are the vertices

so ({{{-2}}},{{{-3+4}}}) = ({{{-2}}},{{{1}}})
and ({{{-2}}}, {{{-3-4}}}) = ({{{-2}}},{{{-7}}})

you can do the same for the minor axis, just moving horizontally up and down by {{{2}}}

{{{drawing( 600, 600, -10, 10, -10, 10, 
         grid(1),
         
         red( ellipse( -2, -3, 4, 8 ) )
          )
)}}}