In the ellipse 25x² + 9y² – 150x – 90y + 225 = 0

$$\begin{gathered} Given ellipse 25x^2+9y^2-150x-90y+225=0 \\ \Rightarrow 25(x^2-6x)+9(y^2-10y)+225=0 \\ \Rightarrow 25{(x-3)}^2+9{(y-5)}^2=225 \Rightarrow \frac{(\mathrm{x}-3{)}^2}{9}+\frac{(\mathrm{y}-5{)}^2}{25}=1(a<b) \\ where a^2=9,b^2=25,centre\left(C\right)=(3,5) \\ and \mathrm{e}=\frac{4}{5} \\ Now \mathrm{be}=4 \\ foci are (h,k\pm be)=(3,5\pm 4) \end{gathered}$$
S(3,1) ; S¹(3,9)