An ellipse with centre (0,0), foci are $(0,5\sqrt{3})$. If the difference of the lengths of major axis and minor axis is 10. Then the length of latusrectum is k then $\frac{25\mathrm{k}}{13}=$…..

Given foci $(0,5\sqrt{3})$ 

Here axis is along y-axis 

$Equation of ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a<b)$
$\begin{array}{l}\Rightarrow \text{ be }=5\sqrt{3}\text{ and }2\mathrm{b}-2\mathrm{a}=10\\ \Rightarrow \mathrm{b}-\mathrm{a}=5\to \left(1\right)\\ Now e^2=1-\frac{a^2}{b^2}\\ \Rightarrow b^2{e}^2=(b-a)(b+a)\\ \Rightarrow 75=5(b+a)\\ (b+a)=15\to \left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow \mathrm{b}=10,\mathrm{a}=5\\ \therefore \text{ length of latusrectum }=\frac{2{\mathrm{a}}^2}{\mathrm{b}}\\ \Rightarrow k =5\\ Now \frac{25\mathrm{k}}{13}=\frac{125}{13}=9.615\end{array}$