JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Ellipse

The focus at $(0,5\sqrt{3})$ lies on the y-axis, so the major axis is vertical and the larger semi-axis is $b$ with $be=5\sqrt{3}$, hence $b^2{e}^2=75$. The axis-difference condition $2b-2a=10$ means $b-a=5$. Using $b^2=a^2+b^2{e}^2=a^2+75$, we get $b^2-a^2=75$, so $(b-a)(b+a)=75$ and $b+a=15$. Solving the pair $b-a=5$, $b+a=15$ gives $b=10$, $a=5$. The latus rectum is $\frac{2a^2}{b}=\frac{2(25)}{10}=5$.