question_answer

For each point\[(x,\text{ }y)\]on an ellipse, the sum of the distances from\[(x,\text{ }y)\]to the points\[(2,0)\]and\[(-2,0)\]is 8. Then, the positive value of\[x\]so that\[(x,3)\]lies on the ellipse is

A)  \[2\]                                    

B)  \[2\sqrt{3}\]

C)  \[\frac{1}{\sqrt{3}}\]                    

D)         \[4\]

E)  \[0\]

Correct Answer: A

Solution :

Given that, \[PS+PS=8\] \[PS=8-PS\] Squaring on both sides, \[{{(PS)}^{2}}=64+{{(PS)}^{2}}-16(PS)\] \[\{{{(x-2)}^{2}}+{{y}^{2}}\}=64+\{{{(x+2)}^{2}}+{{y}^{2}}\}\]                                 \[-16\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}\]                                 \[[\because PS=\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}]\]                                 \[[\because PS=\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}]\] \[({{x}^{2}}+{{y}^{2}}+4-4x)=({{x}^{2}}+{{y}^{2}}+4+4x+64)\] \[-16\sqrt{({{x}^{2}}+{{y}^{2}}+4x+4)}\] \[\Rightarrow \]               \[16\sqrt{({{x}^{2}}+{{y}^{2}}+4+4x)}=8x+64\] \[\Rightarrow \]               \[2\sqrt{{{x}^{2}}+{{y}^{2}}+4x+4}=(x+8)\] Squaring on both sides; \[4{{x}^{2}}+4{{y}^{2}}+16x+16={{x}^{2}}+64+16x\] \[\Rightarrow \]\[3{{x}^{2}}+4{{y}^{2}}=48,\]which is equation of an ellipse given\[(y=3),\because (x,3)\]lies on an ellipse. \[\Rightarrow \]               \[3{{x}^{2}}+4\times 9=48\] \[\Rightarrow \]               \[3{{x}^{2}}=12\] \[\Rightarrow \]               \[{{x}^{2}}=4\]          (here we take only positive value of\[x\].) \[\Rightarrow \]               \[x=\pm 2\] \[\Rightarrow \]               \[x=2\]