If $\mathrm{y}=\mathrm{x}+\mathrm{c}$ is a normal to the ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1$ then ${\mathrm{c}}^2=$

Given ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1$

where  ${\mathrm{a}}^2=25, {\mathrm{b}}^2=9$

Given line $\mathrm{y}=\mathrm{x}+\mathrm{c} \Rightarrow \mathrm{x}-\mathrm{y}+\mathrm{c}=0$

If this is a normal then

$\frac{{\mathrm{a}}^2}{{\mathrm{l}}^2}+\frac{{\mathrm{b}}^2}{{\mathrm{m}}^2}=\frac{{({\mathrm{a}}^2-{\mathrm{b}}^2)}^2}{{\mathrm{n}}^2}$

where $\mathrm{l}=1, \mathrm{m}=-1, \mathrm{n}=\mathrm{c}$

$$\begin{gathered} \Rightarrow \frac{25}{1}+\frac{9}{1}=\frac{{\left(16\right)}^2}{{\mathrm{c}}^2} \\ \Rightarrow 34=\frac{256}{{\mathrm{c}}^2} \\ \Rightarrow {\mathrm{c}}^2=\frac{256}{34} \\ \Rightarrow {\mathrm{c}}^2=\frac{128}{17} \end{gathered}$$