Question 1033118
{{{10x^2 + 80x + 7y^2 + 42y + 83 = 0 }}}

<==> {{{10(x^2+8x+16)+7(y^2+6y+9) = -83+160+63}}}

==> {{{10(x+4)^2+7(y+3)^2 = 140}}}

==> {{{(x+4)^2/14+(y+3)^2/20 = 1}}}

==> the center of the ellipse is the point (-4,-3).

Now {{{a^2 = 20}}} and {{{b^2 = 14}}}  ==> {{{c^2 = 20-14 = 6}}}, because for any ellipse, {{{a^2 = b^2+c^2}}}.
(Also the major axis is parallel to the y-axis, while the minor axis is parallel to the x-axis.)

==> the vertices are (-4, {{{-3+2sqrt(5)}}}) and (-4, {{{-3-2sqrt(5)}}}),
the co-vertices are ({{{-4+sqrt(14)}}},-3) and ({{{-4-sqrt(14)}}},-3), 
while the foci are (-4, {{{-3+sqrt(6)}}}) and (-4, {{{-3-sqrt(6)}}}).