Any tangent on an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by
$$
y=m x \pm \sqrt{a^2 m^2+b^2}
$$
Here $a=2, b=1$
$$
\begin{aligned}
& m=\frac{1-0}{0-2}=-\frac{1}{2} \\
& c=\sqrt{4\left(-\frac{1}{2}\right)^2+1^2}=\sqrt{2}
\end{aligned}
$$
So, $y=-\frac{1}{2} x \pm \sqrt{2}$
For ellipse : $\frac{x^2}{4}+\frac{y^2}{1}=1$
We put $y=-\frac{1}{2} x+\sqrt{2}$

$$
\begin{aligned}
& \therefore \frac{x^2}{4}+\left(-\frac{x}{2}+\sqrt{2}\right)^2=1 \\
& \frac{x^2}{4}+\left(\frac{x^2}{4}-2\left(\frac{x}{2}\right) \sqrt{2}+2\right)=1 \\
& \Rightarrow x^2+2 \sqrt{2} x+2=0 \\
& \text { or } x^2-2 \sqrt{2} x+2=0 \\
& \Rightarrow x=\sqrt{2} \text { or }-\sqrt{2}
\end{aligned}
$$
If $x=\sqrt{2}, y=\frac{1}{\sqrt{2}}$ and $x=-\sqrt{2}, y=-\frac{1}{\sqrt{2}}$
$\therefore$ Points are $\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right),\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$
$$
\left.\therefore P_1 P_2=\sqrt{\left\{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\right\}^2+\left\{\sqrt{2}-(-\sqrt{2}\}^2\right.}\right)
$$

$$
=\sqrt{\left(\frac{2}{\sqrt{2}}\right)^2+\left(2 \sqrt{2}^2\right.}=\sqrt{2+8}=\sqrt{10}
$$