The line 2x + y = 3 cuts the ellipse $4{\mathrm{x}}^2+{\mathrm{y}}^2=5$ at P and Q. If $\mathrm{\theta }$ be the angle between the normals at these points, then $5\tan \theta =$

Solving $4{\mathrm{x}}^2+{\mathrm{y}}^2=5\text{ with the line }\mathrm{y}=3-2\mathrm{x}$ we get the point of intersection as $\mathrm{P}\left(\frac{1}{2},2\right)$ and Q(1,1)
Tangent at $\left(\frac{1}{2},2\right)\text{ is }4\mathrm{x}\left(\frac{1}{2}\right)+\mathrm{y}\cdot 2=5$, slope is -1
Tangent at (1,1) is 4x.(1) + y.1= 5 slope is -4
Slopes of the normals are 1,1 / 4 . If $\mathrm{\theta }$ be the angle between the normals, then
$\tan \mathrm{\theta }=\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}=\frac{1-1/4}{1+1/4}=\frac{3}{5}$