If  $a{x}^2+b{y}^2=15$  is the equation of the ellipse for which distance between its foci is 2 and distance between its directrices is 5, then  a+b=

$$\begin{gathered} a{x}^2+b{y}^2=15\Rightarrow \frac{x^2}{{\left(\sqrt{\frac{15}{a}}\right)}^2}+\frac{y^2}{{\left(\sqrt{\frac{15}{b}}\right)}^2}=1 \\ {a}^1=\sqrt{\frac{15}{a}},b^1=\sqrt{\frac{15}{b}} \\ \Rightarrow 2a^{1}e=2,\frac{2a^1}{e}=5 \\ \Rightarrow a^{1}e=1\frac{a^1}{e}=\frac{5}{2} \\ \begin{array}{l}\Rightarrow a^1=\pm \frac{\sqrt{5}}{\sqrt{2}}\Rightarrow e=\pm \frac{\sqrt{2}}{\sqrt{5}}\\ \Rightarrow b^{12}=a^{12}(1-e^2)=\frac{5}{2}(1-\frac{2}{5})=\frac{3}{2}\end{array} \\ {b}^1=\pm \frac{\sqrt{3}}{\sqrt{2}} \\ \therefore \frac{\sqrt{5}}{\sqrt{2}}=\sqrt{\frac{15}{a}},\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{15}{b}} \\ \Rightarrow a=6,b=10 \\ a+b=16 \end{gathered}$$