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(B) The given curve is $y^{2} = \left|\begin{array}{ll}x+1 & x+2 \ x+3 & x+5\end{array}ight|$.Expanding the determinant:$y^{2} = (x+1)(x+5) - (x+2)

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$x$-axis

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(B) The given curve is $y^{2} = \left|\begin{array}{ll}x+1 & x+2 \ x+3 & x+5\end{array}ight|$.Expanding the determinant:$y^{2} = (x+1)(x+5) - (x+2)(x+3)$$y^{2} = (x^{2} + 6x + 5) - (x^{2} + 5x + 6)$$y^{2} = x - 1$,which is a parabola.Let the parametric coordinates of a point $Q$ on the parabola be $Q(t^{2}+1, t)$.Given $P(x, y)$ is the midpoint of the line segment joining $A(1, 0)$ and $Q(t^{2}+1, t)$:$x = \frac{1 + t^{2} + 1}{2} = \frac{t^{2} + 2}{2} \Rightarrow t^{2} = 2x - 2$$y = \frac{0 + t}{2} = \frac{t}{2} \Rightarrow t = 2y$Substituting $t$ in the equation for $x$:$(2y)^{2} = 2x - 2$$4y^{2} = 2(x - 1)$$y^{2} = \frac{1}{2}(x - 1)$This is a parabola with its axis along the $x$-axis. Therefore,the locus of $P$ is symmetrical about the $x$-axis.

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

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