Question 681711
Vertical Ellipse


{{{(x-h)^2/b^2+(y-k)^2/a^2}}}

{{{ 4x^2+(y-2)^2=16 }}}.......both sides divide by {{{16}}}

{{{ 4x^2/16+(y-2)^2/16=16/16 }}}

{{{ x^2/4+(y-2)^2/16=16}}}, so you have


{{{ (x-0)^2/4+(y-2)^2/16=16}}}

{{{b=2}}}, {{{a=4}}}, {{{h=0}}}, and {{{k=2}}}


Center: ({{{h}}}, {{{k}}})=({{{0}}}, {{{2}}})

Vertices: ({{{0}}}, {{{2}}} ± {{{4}}})  or

        ({{{0}}},{{{4}}}), ({{{0}}},{{{-4}}})

Co-vertices: ({{{0}}} ± {{{2}}},{{{ 2}}}) or ({{{2}}},{{{0}}}), and ({{{-2}}},{{{0}}})

Length of Major Axis: {{{2a=8}}}

Length of Minor Axis: {{{2b=4}}}



{{{graph( 600, 600, -10, 10, -10, 10,(x-0)^2/4+(y-2)^2/16>=1 )}}}


{{{drawing(600,600,-5,10,-5,10,grid(0),circle(0,2,0.2),graph(600,600,-5,10,-5,10,sqrt(-4x^2+14)+2,-sqrt(-4x^2+14)+2))}}}