Any tangent on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ given by

$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $

Here $a = 2,b = 1$

$m = \frac{{1 – 0}}{{0 – 2}} = \frac{1}{2}$

$c = \sqrt {4{{\left( { – \frac{1}{2}} \right)}^2} + {1^2}}  = \sqrt 2 $

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 \pm \sqrt 2 $

$\therefore \frac{{{x^2}}}{4} + {\left( { – \frac{x}{2} + \sqrt 2 } \right)^2} = 1$

$\frac{{{x^2}}}{4} + \left( {\frac{{{x^2}}}{4} – 2\left( {\frac{x}{2}} \right)\sqrt 2  + 2} \right) = 1$

$ \Rightarrow {x^2} + 2\sqrt 2 x + 2 = 0$

or ${x^2} – 2\sqrt 2 x + 2 = 0$

$ \Rightarrow x = \sqrt 2 $ or $ – \sqrt 2 $

If $x = \sqrt 2 ,y = \frac{1}{{\sqrt 2 }}$ and $x =  – \sqrt 2 ,y =  – \frac{1}{{\sqrt 2 }}\,$

$\therefore $ Points are $\left( {\sqrt 2 ,\frac{1}{{\sqrt 2 }}} \right),\left( { – \sqrt 2 , – \frac{1}{{\sqrt 2 }}} \right)\,$

$\therefore {P_1}{P_2} = \sqrt {{{\left\{ {\frac{1}{{\sqrt 2 }} – \left( {\frac{1}{{\sqrt 2 }}} \right)} \right\}}^2} + {{\left\{ {\sqrt 2  – \left( { – \sqrt 2 } \right)} \right\}}^2}} \,$

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