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If $A=(2,3,4)$ and $B=(-2,3,4)$, then the locus of a point $P$ such that $\mathrm{PA}+\mathrm{PB}=4$ is
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The correct answer is:
$y^2+z^2-6 y-8 z+25=0$
Given $A=(2,3,4), B=(-2,3,4)$
$\begin{aligned}
& \text { Now, PA }+P B=4 \Rightarrow P A^2=(4-P B)^2 \\
& \Rightarrow(x-2)^2+(y-3)^2+(z-4)^2=16+P^2-8 P B \\
& \Rightarrow(x-2)^2+(y-3)^2+(z-4)^2 \\
& =16+(z+2)^2+(y-3)^2+(z-4)^2-8 P B \\
& \Rightarrow(x-2)^2=16+(x+2)^2-8 P B \\
& \Rightarrow 8 x+16=8 P B \Rightarrow 64 P^2=(8 x+16)^2 \\
& \Rightarrow 64\left((x+2)^2+(y-3)^2+(z-4)^2\right) \\
& =64 x^2+256+256 x \\
& \Rightarrow 64 x^2+64 y^2+64 z^2+256 x-384 y-512 z+1856 \\
& =64 x^2+256+256 x \\
& \Rightarrow 64 y^2+64 z^2-384 y-518 z+1856-256=0 \\
& \Rightarrow y^2+z^2-6 y-8 z+25=0
\end{aligned}$
$\begin{aligned}
& \text { Now, PA }+P B=4 \Rightarrow P A^2=(4-P B)^2 \\
& \Rightarrow(x-2)^2+(y-3)^2+(z-4)^2=16+P^2-8 P B \\
& \Rightarrow(x-2)^2+(y-3)^2+(z-4)^2 \\
& =16+(z+2)^2+(y-3)^2+(z-4)^2-8 P B \\
& \Rightarrow(x-2)^2=16+(x+2)^2-8 P B \\
& \Rightarrow 8 x+16=8 P B \Rightarrow 64 P^2=(8 x+16)^2 \\
& \Rightarrow 64\left((x+2)^2+(y-3)^2+(z-4)^2\right) \\
& =64 x^2+256+256 x \\
& \Rightarrow 64 x^2+64 y^2+64 z^2+256 x-384 y-512 z+1856 \\
& =64 x^2+256+256 x \\
& \Rightarrow 64 y^2+64 z^2-384 y-518 z+1856-256=0 \\
& \Rightarrow y^2+z^2-6 y-8 z+25=0
\end{aligned}$
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