If $\mathrm{x}+\mathrm{y}+\mathrm{n}=0, \mathrm{n}>0$ is a normal to the ellipse ${\mathrm{x}}^2+3{\mathrm{y}}^2=3$ and $\mathrm{x}+\mathrm{my}+3=0$ and $\mathrm{m}<0$ is a tangent to the ellipse ${\mathrm{x}}^2+5{\mathrm{y}}^2=5$ then the point of intersection of  these two lines satisfies the equation

Given ellipses are 

$\frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{1}=1 -\left(1\right)$

and $\frac{{\mathrm{x}}^2}{5}+\frac{{\mathrm{y}}^2}{1}=1 -\left(2\right)$

Given lines are $\mathrm{y}=\frac{-\mathrm{x}}{\mathrm{m}}-\frac{3}{\mathrm{m}} -\left(3\right)$ 

and $\mathrm{x}+\mathrm{y}+\mathrm{n}=0 -\left(4\right)$

since (3) is tangent to (2) then 

$$\begin{gathered} \Rightarrow \frac{9}{{\mathrm{m}}^2}=5\left(\frac{1}{{\mathrm{m}}^2}\right)+1 \\ \Rightarrow \frac{4}{{\mathrm{m}}^2}=1 \\ \Rightarrow {\mathrm{m}}^2=4 \\ \Rightarrow \mathrm{m}=-2 (\because \mathrm{m}<0) \end{gathered}$$

since (4) is normal to (1) then 

$\begin{array}{rcl} & \Rightarrow & n=\pm 1 \Rightarrow \mathrm{n}=1(\because \mathrm{n}>0)\\ \left(3\right) & \Rightarrow & \mathrm{x}-2\mathrm{y}+3=0\\ \left(4\right) & \Rightarrow & \mathrm{x}+\mathrm{y}+1=0\end{array}$

By solving these two equations we get $point of intersection=(\frac{-5}{3},\frac{2}{3})$

By verification it satisfies  the line x-5y +5=0