Q84 / 225 · JEE Main 2005 · Mathematics

Let $P$ be the point $(1,0)$ and $Q$ a point on the locus $y^2=8 x$. The locus of mid point of $P Q$ is

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

Solution

$\mathrm{P}=(1,0)$ $Q=(h, k)$ such that $k^2=8 h$ Let $(\alpha, \beta)$ be the midpoint of PQ $$ \begin{aligned} & \alpha=\frac{\mathrm{h}+1}{2}, \quad \beta=\frac{\mathrm{k}+0}{2} \\ & 2 \alpha-1=\mathrm{h}, \quad 2 \beta=\mathrm{k} . \\ & (2 \beta)^2=8(2 \alpha-1) \Rightarrow \beta^2=4 \alpha-2 \\ & \Rightarrow \mathrm{y}^2-4 \mathrm{x}+2=0 . \end{aligned} $$

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

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