If the line x + y + k= 0 is a normal  to the hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$ then k=

$$\begin{gathered} \text{We know that if the line lx+my+n=0 is normal to the hyperbola }\frac{x^2}{a^2}\text{-}\frac{y^2}{b^2}\text{=1 then } \\ \frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{{\left(a^2+b^2\right)}^2}{n^2} \\ Given ellipse \frac{x^2}{9}-\frac{y^2}{4}=1 \\ where a^2=9,b^2=4 \\ Given line x+y+k=0\to \left(1\right) \\ \Rightarrow l=1,m=1,n=k \\ \sin ce \left(1\right) is a normal then by applying apply above formula we get \\ 9-4=\frac{{\left(13\right)}^2}{k^2} \\ \Rightarrow k^2=\frac{169}{5} \\ \Rightarrow k=\pm \frac{13}{\sqrt{5}} \end{gathered}$$