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  • For each point on an ellipse, the sum of the distances from <img alt="\mathrm {

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

Option: 1

2


Option: 2

3


Option: 3

6


Option: 4

4


Answers (1)

best_answer

Now, \mathrm {\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=8} 


Squaring both sides


\begin{aligned} & \mathrm {(x-2)^2+y^2=64+(x+2)^2+y^2-2 \times 8 \sqrt{(x+2)^2+y^2}} \\ &\mathrm { \Rightarrow \quad-4 x=64+4 x-16 \sqrt{(x+2)^2+y^2}} \\ &\mathrm { \Rightarrow \quad-x-8=-2 \sqrt{(x+2)^2+y^2} }\\ & \mathrm {\text { put } y=3 \text { in (ii), we get }} \\ & \mathrm {\quad-x-8=-2 \sqrt{(x+2)^2+9}} \end{aligned}

put  \mathrm y=3  in (ii), we get


\mathrm { -x-8=-2 \sqrt{(x+2)^2+9}} 

Squaring both sides

{ \begin{aligned} & \mathrm {x^2+64+16 x=4(x+2)^2+36 \Rightarrow 3 x^2=12 }\\ & \mathrm {x^2=4 \Rightarrow x= \pm 2} \end{aligned} }

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Rishabh

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