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

Solving $4x^2+y^2=5\text{ with the line }y=3-2x$ we get the point of intersection as $P\left(\frac{1}{2},2\right)\text{ and }Q(1,1)$ 

Tangent at $\left(\frac{1}{2},2\right)\text{ is }4x\cdot \left(\frac{1}{2}\right)+y2=5$ slope is -1  and slope of tangent at Q(1,1) is -4

$\therefore$Slopes of the normal  are 1, 1/4. If  $\theta$ be the angle  between the normal, then  

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{1-1/4}{1+1/4}=\frac{3}{5}$