Let $P(x, y)$ be the mid point of the line joining $(1,0)$ to a point on the curve $\mathrm{y}^{2}=\left|\begin{array}{ll}\mathrm{x}+1 & \mathrm{x}+2 \\ \mathrm{x}+3 & \mathrm{x}+5\end{array}\right|$. Then, locus of $\mathrm{P}$ is symmetrical about

• A

  $y$-axis
• Correct answer
  B

  $x$-axis
• C

  $\mathrm{x}=1$
• D

  $y=1$

Solution

The given curve $\mathrm{y}^{2}=\left|\begin{array}{ll}\mathrm{x}+1 & \mathrm{x}+2 \\ \mathrm{x}+3 & \mathrm{x}+5\end{array}\right|$

$$ \begin{aligned} &\Rightarrow \quad \mathrm{y}^{2}=(\mathrm{x}+1)(\mathrm{x}+5)-(\mathrm{x}+2)(\mathrm{x}+3) \\ &\Rightarrow \quad \mathrm{y}^{2}=\mathrm{x}^{2}+6 \mathrm{x}+5-\mathrm{x}^{2}-5 \mathrm{x}-6 \\ &\quad \mathrm{y}^{2}=\mathrm{x}-1 \text { which is parabola. } \end{aligned} $$ Let the parametric coordinate of parabola is $$ \mathrm{Q}=\left(\frac{\mathrm{t}^{2}}{4}, \frac{\mathrm{t}}{2}+1\right) $$ Now, given that $\mathrm{P}(\mathrm{x}, \mathrm{y})$ is the mid point of $\mathrm{AQ}$. $$ P(x, y)=\left[\frac{\frac{t^{2}}{4}+1}{2}, \frac{\frac{t}{2}+1}{2}\right] $$ $$ \begin{array}{ll} \Rightarrow & \mathrm{x}=\frac{\mathrm{t}^{2}}{8}+\frac{1}{2} \Rightarrow 8 \mathrm{x}=\mathrm{t}^{2}+4 \\ \text { and } & \mathrm{y}=\frac{\mathrm{t}}{4}+\frac{1}{2} \Rightarrow \mathrm{t}=2(2 \mathrm{y}-1) \\ \Rightarrow & 8 \mathrm{x}=4(2 \mathrm{y}-1)^{2}+4 \\ \Rightarrow & 2 \mathrm{x}=(2 \mathrm{y}-1)^{2}+1 \\ \Rightarrow & (2 \mathrm{y}-1)^{2}=(2 \mathrm{x}-1) \\ \Rightarrow & 4\left(\mathrm{y}-\frac{1}{2}\right)^{2}=2\left(\mathrm{x}-\frac{1}{2}\right) \\ \Rightarrow & \left(\mathrm{y}-\frac{1}{2}\right)^{2}=\frac{1}{2}\left(\mathrm{x}-\frac{1}{2}\right) \end{array} $$ Which is also a parabola. Hence, the locus of $\mathrm{P}(\mathrm{x}, \mathrm{y})$ is symmetrical about $x$-axis ie, $\mathrm{y}=0$ from figure.

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