Let $A$ be the point $(0,4)$ and $B$ be a moving point on $x$-axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$-axis at $R$. The locus of the midpoint $P$ of $M R$ is

• Correct answer
  A

  $y+x^{2}=2$
• B

  $x^{2}+(y-2)^{2}=\frac{1}{4}$
• C

  $(y-2)^{2}-x^{2}=\frac{1}{4}$
• D

  $x^{2}+y^{2}=16$

Solution

Let B $=(2 \alpha, 0)$ $\therefore \mathrm{M}=(\alpha, 2)$ $\mathrm{m}_{\mathrm{AB}}=\frac{4}{-2 \alpha}=-\frac{2}{\alpha}$ $\therefore \mathrm{m}_{\mathrm{MR}}=\frac{\alpha}{2}, \therefore \mathrm{MR} \quad \mathrm{y}-2=\frac{\alpha}{2}(\mathrm{x}-\alpha)$ Put, $\mathrm{x}=0 \Rightarrow \mathrm{y}=2-\frac{\alpha^{2}}{2}$ $\therefore \mathrm{R} \equiv\left(0,2-\frac{\alpha^{2}}{2}\right)$ $\therefore \mathrm{P} \equiv\left(\frac{\alpha}{2}, \frac{2-\alpha^{2} / 2+2}{2}\right) \equiv \left(\frac{\alpha}{2}, 2-\frac{\alpha^{2}}{4}\right)$ $\therefore \mathrm{y}=2-\mathrm{x}^{2} \Rightarrow \mathrm{y}+\mathrm{x}^{2}=2$

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