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If $A(2,3)$ and $B(2,-3)$ are two points, then the equation of the locus of a point $\mathrm{P}$ such that $\mathrm{PA}+\mathrm{PB}=8$ is
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The correct answer is:
$16 x^2+7 y^2-64 x-48=0$
Let co-ordinate of point $\mathrm{P}$ is $(\mathrm{x}, \mathrm{y})$
Now, $\mathrm{PA}+\mathrm{PB}=8$
$\begin{aligned} & \Rightarrow \sqrt{(x-2)^2+(y-3)^2}+\sqrt{(x-2)^2+(y+3)^2}=8 \\ & \Rightarrow \sqrt{(x-2)^2+(y-3)^2}=8-\sqrt{(x-2)^2+(y+3)^2} \\ & \Rightarrow(x-2)^2+(y-3)^2=64+(x-2)^2+(y+3)^2 \\ & -16 \sqrt{(x-2)^2+(y+3)^2} \\ & \Rightarrow 4 \sqrt{(x-2)^2+(y+3)^2}=16+3 y \\ & \Rightarrow 16\left(x^2+4-4 x+y^2+9+6 y\right)=256+9 y^2+96 y \\ & \Rightarrow 16 x^2+7 y^2-64 x-48=0\end{aligned}$
Now, $\mathrm{PA}+\mathrm{PB}=8$
$\begin{aligned} & \Rightarrow \sqrt{(x-2)^2+(y-3)^2}+\sqrt{(x-2)^2+(y+3)^2}=8 \\ & \Rightarrow \sqrt{(x-2)^2+(y-3)^2}=8-\sqrt{(x-2)^2+(y+3)^2} \\ & \Rightarrow(x-2)^2+(y-3)^2=64+(x-2)^2+(y+3)^2 \\ & -16 \sqrt{(x-2)^2+(y+3)^2} \\ & \Rightarrow 4 \sqrt{(x-2)^2+(y+3)^2}=16+3 y \\ & \Rightarrow 16\left(x^2+4-4 x+y^2+9+6 y\right)=256+9 y^2+96 y \\ & \Rightarrow 16 x^2+7 y^2-64 x-48=0\end{aligned}$
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