If P_1 and P_2 are two points on the ellipse (x^2)/(4)+y^2=1 at...

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$$\begin{aligned}& m=\frac{1-0}{0-2}=-\frac{1}{2}\\ & c=\sqrt{4\left(-\frac{1}{2}\right)^2+1^2}=\sqrt{2} \end{aligned}$So, $y=-\frac{1}{2} x \pm \sqrt{2}$ For ellipse : $\frac{x^2}{4}+\frac{y^2}{1}=1$ We put $y=-\frac{1}{2} x+\sqrt{2}\begin{aligned}& \therefore \frac{x^2}{4}+\left(-\frac{x}{2}+\sqrt{2}\right)^2=1 \\ & \frac{x^2}{4}+\left(\frac{x^2}{4}-2\left(\frac{x}{2}\right) \sqrt{2}+2\right)=1 \\ & \Rightarrow x²+2 \sqrt{2}x+2=0 \\ & \text{ or }x^2-2 \sqrt{2}x+2=0 \\ & \Rightarrow x=\sqrt{2} \text{ or }-\sqrt{2} \end{aligned}$$$If$x=\sqrt{2}, y=\frac{1}{\sqrt{2}}$and$x=-\sqrt{2}, y=-\frac{1}{\sqrt{2}}\therefore$Points are$\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right),\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)\left.\thereforeP_1 P_2=$\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}