A tangent to the ellipse ${\mathrm{x}}^2+4{\mathrm{y}}^2=4$ meets the ellipse ${\mathrm{x}}^2+2{\mathrm{y}}^2=6$ at P and Q. The angle between the tangents at P and Q of the ellipse ${\mathrm{x}}^2+2{\mathrm{y}}^2=6$ is

Given ellipses are ${\mathrm{x}}^2+2{\mathrm{y}}^2=6 -\left(1\right)$

where ${\mathrm{a}}^2=6, {\mathrm{b}}^2=3$

Let R(h, k) be the point of intersection of tangents at P and Q of (1)

Now $\overline{\mathrm{PQ}}$ is chord of contact of (1) 

Now equation of chord of contact is ${\mathrm{S}}_1=0$

$$\begin{gathered} \Rightarrow \mathrm{x}\left(\mathrm{h}\right)+2\mathrm{y}\left(\mathrm{k}\right)-6=0 \\ \Rightarrow \mathrm{h} \mathrm{x}+2\mathrm{yk}-6=0 -\left(2\right) \end{gathered}$$

This line is tangent to the ellipse ${\mathrm{x}}^2+4{\mathrm{y}}^2=4 -\left(3\right)$

Let $\mathrm{P}=(2\mathrm{cos\theta }, \mathrm{sin\theta })$ be any point on (3).

Now equation of tangent at P to the ellipse (3) is $2\mathrm{xcos\theta }+4\mathrm{y} \mathrm{sin\theta }-4=0 -\left(4\right)$

since (2) and (4) represents the same line then 

$$\begin{gathered} \frac{\mathrm{h}}{2\mathrm{cos\theta }}=\frac{2\mathrm{k}}{4\mathrm{sin\theta }}=\frac{\cancel{-}6}{\cancel{-}4} \\ \Rightarrow \mathrm{h}=3\mathrm{cos\theta }, \mathrm{k}=3\mathrm{sin\theta } \end{gathered}$$

Now ${\mathrm{h}}^2+{\mathrm{k}}^2=9({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })=9\left(1\right)=9$

$\Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2=6+3={\mathrm{a}}^2+{\mathrm{b}}^2$

$\therefore$ R lies on director circle.

$\therefore$ Angle $\angle \mathrm{PRQ}={90}^0$