Question 1204200
The standard form of the equation of an ellipse with center 
({{{ h}}}, {{{ k}}}) and major axis parallel to the {{{ x}}}-axis is

{{{ (x-h)^2/a^2+(y-k)^2/b^2=1}}}

where
{{{ a>b}}}
the length of the major axis is {{{ 2a}}}
the coordinates of the vertices are ({{{  h }}}±{{{a}}}, {{{ k}}})
the length of the minor axis is{{{  2b}}}
the coordinates of the co-vertices are ({{{  h}}},{{{  k}}} ±{{{b}}})
the coordinates of the foci are ({{{  h}}} ± {{{c}}}, {{{ k}}}), where {{{c^2=a^2-b^2}}}

the ellipse that has 

a center at ({{{ 5}}},{{{ -4}}}) 
 
Thus, {{{ h=5}}}, {{{ k=-4}}}

vertex at ({{{ 12}}},{{{ -4}}})=> ({{{ 5}}} ± {{{ a}}},{{{ -4}}})=>

{{{ 5+a=12}}}
 {{{ a=12-5}}}
{{{ a=7}}}

a covertex at ({{{ 5}}},{{{ -6}}})=>({{{ 5}}},{{{ -4}}}±{{{ b}}})
{{{ -4+b=-6}}}
{{{ b=6-4}}}
{{{ b=2}}}

your ellipse is

{{{ (x-5)^2/7^2+(y-(-4))^2/2^2=1}}}

{{{ (x-5)^2/49+(y+4)^2/4=1}}}



{{{ drawing( 600, 600, -10, 15, -10, 10,
circle(5,-4,.12), locate(5,-4,C(5,-4)),
circle(12,-4,.12), locate(12,-4,V(12,-4)),
circle(5,-6,.12), locate(5,-6,co-V(5,-6)),
graph( 600, 600, -10, 15, -10, 10, (2/7)(-sqrt(-x^2+10x+24)-14), (2/7)(sqrt(-x^2+10x+24)-14))) }}}