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

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
  1. $y^2+z^2+6 y+8 z+25=0$
  2. $y^2-z^2+6 y+8 z-25=0$
  3. $y^2+z^2-6 y-8 z+25=0$
  4. $y^2+z^2-6 y-8 z-25=0$

Solution

Given $A=(2,3,4), B=(-2,3,4)$ $\begin{aligned} & \text { Now, PA }+P B=4 \Rightarrow P A^2=(4-P B)^2 \\ & \Rightarrow(x-2)^2+(y-3)^2+(z-4)^2=16+P^2-8 P B \\ & \Rightarrow(x-2)^2+(y-3)^2+(z-4)^2 \\ & =16+(z+2)^2+(y-3)^2+(z-4)^2-8 P B \\ & \Rightarrow(x-2)^2=16+(x+2)^2-8 P B \\ & \Rightarrow 8 x+16=8 P B \Rightarrow 64 P^2=(8 x+16)^2 \\ & \Rightarrow 64\left((x+2)^2+(y-3)^2+(z-4)^2\right) \\ & =64 x^2+256+256 x \\ & \Rightarrow 64 x^2+64 y^2+64 z^2+256 x-384 y-512 z+1856 \\ & =64 x^2+256+256 x \\ & \Rightarrow 64 y^2+64 z^2-384 y-518 z+1856-256=0 \\ & \Rightarrow y^2+z^2-6 y-8 z+25=0 \end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

Practice more Straight Lines questions on Aicharya