The length of the latusrectum of $9{\mathrm{x}}^2 + 25{\mathrm{y}}^2 – 90\mathrm{x} – 150\mathrm{y} + 225 = 0$ is

$$\begin{gathered} 9({\mathrm{x}}^2-10+25-25)+25({\mathrm{y}}^2-6\mathrm{y}+9-9)+225=0 \\ 9{(\mathrm{x}-5)}^2-225+25{(\mathrm{y}-3)}^2-225+225=0 \\ 9{(\mathrm{x}-5)}^2+25{(\mathrm{y}-3)}^2=225 \\ \Rightarrow \frac{{(\mathrm{x}-5)}^2}{25}+\frac{{(\mathrm{y}-3)}^2}{9}=1 \\ where a^2=25 ,b^2=9 \\ Now L.L.R=\frac{2b^2}{a}=\frac{18}{5} \end{gathered}$$