Any tangent on an ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { is given by }$

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

$\text { Here } a=2, b=1$

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

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

$\text { so, } y=-\frac{1}{2} x \pm \sqrt{2}$

$\text { for ellipse }: \frac{x^2}{4}+\frac{y^2}{1}=1$

$\text { we put } y=-\frac{1}{2} x+\sqrt{2}$

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

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

$\Rightarrow x^2+2 \sqrt{2} x+2=0(\text { or })$

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

$\Rightarrow x=\sqrt{2}(\text { or }) x=-\sqrt{2}$

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

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

$\therefore \text { P1 } p_2=\sqrt{\left[\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\right]^2+\left[\sqrt{2}-(-\sqrt{2})^2\right]}$

$=\sqrt{10}$