A chord AB of an ellipse $\frac{x^2}{9}+\frac{y^2}{{\lambda }^2}=1$ subtends a right angle at the centre of ellipse O
(where O is origin), also $\frac{1}{{\mathrm{OA}}^2}+\frac{1}{{\mathrm{OB}}^2}=\frac{14}{45},\text{ then }{\lambda }^2$is equal to _____

Let $\mathrm{OA}={\mathrm{r}}_1\text{ and }\mathrm{OB}={\mathrm{r}}_2$

$\Rightarrow A=\left(r_1\cos \theta ,r_1\sin \theta \right)$
$A lies on ellipse$ $\frac{x^2}{9}+\frac{y^2}{{\lambda }^2}=1$
$\begin{array}{r}\Rightarrow \frac{{\mathrm{r}}_1^2{\cos }^2\theta }{9}+\frac{{\mathrm{r}}_1^2{\sin }^2\theta }{{\lambda }^2}=1\\ \Rightarrow \frac{1}{{\mathrm{r}}_1^2}=\frac{{\cos }^2\theta }{9}+\frac{{\sin }^2\theta }{{\lambda }^2} ....i\end{array}$
Now replace $\theta \text{ by }90+{\theta }_1$we get
$\begin{array}{l}\Rightarrow \frac{1}{{\mathrm{r}}_2^2}=\frac{{\sin }^2\theta }{9}+\frac{{\cos }^2\theta }{{\lambda }^2} ....\left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow \frac{14}{45}=\frac{1}{9}+\frac{1}{{\lambda }^2}\\ \Rightarrow \frac{9}{45}=\frac{1}{{\lambda }^2}\Rightarrow {\lambda }^2=5\end{array}$