Q81 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

Let $A=(2,0)$ and $B=(0,-2)$. Let $P$ be any point such that the sum of the distances of $P$ from $A$ and $B$ is 4 . Then the equation of the locus of the point $P$ is

  • A
    $3 x^2-2 x y+3 y^2-4 x+12 y+16=0$
  • Correct answer
    B
    $3 x^2-2 x y+3 y^2-8 x+8 y=0$
  • C
    $3 x^2+2 x y+3 y^2+8 x-8 y=0$
  • D
    $3 x^2+2 x y+3 y^2+4 x-12 y+16=0$

Solution

Let $P(x, y)$ be the point. Hence $P A+P B=4$ $$ \begin{aligned} & \Rightarrow \sqrt{(x-2)^2+(y-0)^2}+\sqrt{(x-0)^2+(y+2)^2}=4 \\ & \Rightarrow \sqrt{(x-2)^2+y^2}=4-\sqrt{x^2+(y+2)^2} \\ & \Rightarrow(x-2)^2+y^2 \\ &=(4)^2+(x)^2+(y+2)^2-8 \sqrt{x^2+(y+2)^2} \\ & \Rightarrow x^2+y^2-4 x+4=16+x^2 \\ & \quad+y^2+4+4 y-8 \sqrt{x^2+(y+2)^2} \\ & \Rightarrow-4 x=16+4 y-8 \sqrt{x^2+(y+2)^2} \\ & \Rightarrow x+y+4=2 \sqrt{x^2+(y+2)^2} \end{aligned} $$ $\begin{aligned} & \Rightarrow \quad x^2+y^2+16+2 x y+8 y+8 x=4\left(x^2+y^2+4+4 y\right) \\ & \Rightarrow \quad 3 x^2-2 x y+3 y^2-8 x+8 y=0 .\end{aligned}$

Hence, the correct answer is option: 'B'.

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