The major axis and minor axis of an ellipse are respectively, x - 2y - 5 - 0 and 2x + y + 10 = 0. If one end of latus rectum is (3 ,4), then the foci are

Foot of perpendicular from one end of latus rectum L(3, 4) on major axis which is focus is given by
$\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-4}{-2}=\frac{-(3-2(4)-5)}{1^2+(-2{)}^2}=2$
$\therefore \mathrm{S}(\mathrm{x},\mathrm{y})\equiv \mathrm{S}(5,0)$
Another focus is image of s in the line 2x + y + 10 = 0.
$\begin{array}{l}\frac{\mathrm{x}-5}{2}=\frac{\mathrm{y}-0}{1}=\frac{-2\left(2\right(5)+0+10)}{2^2+1^2}=-8\\ \therefore {\mathrm{S}}^{\mathrm{\prime }}(\mathrm{x},\mathrm{y})\equiv \mathrm{S}(-11,-8)\end{array}$