Question 1034016
We have to get this into standard form...so from
{{{5x^2+7y^2+10x-28y-2=0}}}
we will some factoring and completing the squares...we get
{{{5(x^2 + 2x) + 7(y^2 - 4y) = 2}}}
{{{5(x^2 + 2x + 1) + 7(y^2 - 4y + 4) = 2 + 5 + 28}}}
{{{5(x+1)^2 + 7(y-2)^2 = 35}}}
{{{(x+1)^2 / (35/5) + (y-2)^2 / (35/7) = 1}}}
Here a^2 = 7 and b^2 = 5...
You need to find 
(c,0) and (-c,0)
where 
c^2 = a^2 - b^2
Take it from here...