Q46 / 155 · WBJEE 2022 · Mathematics

Let $P$ be the point $(2,0)$ and $Q$ be a variable point on $(y-6)^2=2(x-4)$. Then the locus of mid-point of $P Q$ is

  • A
    $y^2+x+6 y+12=0$
  • B
    $y^2-x+6 y+12=0$
  • C
    $y^2+x-6 y+12=0$
  • Correct answer
    D
    $y^2-x-6 y+12=0$

Solution

$$ \begin{gathered} \mathrm{P}(2,0) \\ \mathrm{Q}\left(4+\frac{1}{2} \mathrm{t}^2, 6+\mathrm{t}\right) \end{gathered} $$ Let midpoint of $\mathrm{PQ}$ be R(h, k) $$ \begin{aligned} &\mathrm{h}=3+\frac{1}{4} \mathrm{t}^2, \mathrm{k}=3+\frac{\mathrm{t}}{2} \\ &\Rightarrow(\mathrm{k}-3)^2=(\mathrm{h}-3) \\ &\Rightarrow \mathrm{y}^2-6 \mathrm{y}-\mathrm{x}+12=0 \end{aligned} $$

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

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