A) \[2\sqrt{2}\]
B) \[\sqrt{5}\]
C) \[2\sqrt{3}\]
D) \[\sqrt{10}\]
Correct Answer: D
Solution :
Any tangent on an ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is 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 : \[\frac{{{x}^{2}}}{4}x+\frac{{{y}^{2}}}{1}=1\] We put \[y=-\frac{1}{2}x+\sqrt{2}\] \[\therefore \]\[\frac{{{x}^{2}}}{4}+{{\left( -\frac{x}{2}+\sqrt{2} \right)}^{2}}=1\] \[\frac{{{x}^{2}}}{4}+\left( \frac{{{x}^{2}}}{4}-\left( \frac{x}{2} \right)\sqrt{2}+2 \right)=1\] \[\Rightarrow \]\[{{x}^{2}}+2\sqrt{2}x+2=0\]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\[\left( \sqrt{2},\frac{1}{\sqrt{2}} \right),\left( -\sqrt{2},-\frac{1}{\sqrt{2}} \right)\] \[\therefore \]\[{{P}_{1}}{{P}_{2}}=\sqrt{{{\left\{ \frac{1}{\sqrt{2}}-\left( -\frac{1}{\sqrt{2}} \right) \right\}}^{2}}+{{\left\{ \sqrt{2}-\left( -\sqrt{2} \right) \right\}}^{2}}}\] \[=\sqrt{{{\left( -\frac{1}{\sqrt{2}} \right)}^{2+}}+{{\left( 2\sqrt{2} \right)}^{2}}}=\sqrt{2+8}=\sqrt{10}\]
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