Let $A$ be the point $(0,4)$ in the $xy$-plane and let $B$ be the point $(2t, 0)$. Let $L$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $M$. Let $N$ be the midpoint of $LM$. Then the locus of $N$ is

Let $A=(1, 2)$,$B=(2, 1)$,and $C=(-1, -1)$ be three points. If $P(x, y)$ is a point such that the area of the quadrilateral $PABC$ is twice the area of the triangle $PAB$,then the equation of the locus of $P$ is:

$A$ variable line passes through a fixed point $(x_{1}, y_{1})$ and meets the axes at $A$ and $B$. If the rectangle $OAPB$ is completed,the locus of $P$ is,($O$ being the origin of the system of axes).

$A$ point $P$ on a line is at a distance of $4$ units from the origin $(0,0)$. If the line makes an angle of $60^{\circ}$ with the negative direction of the $X$-axis,then the coordinates of $P$ are

If a point $(x, y) \equiv (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

Starting at time $t=0$ from the origin with speed $1 \text{ m/s}$,a particle follows a two-dimensional trajectory in the $x-y$ plane so that its coordinates are related by the equation $y=\frac{x^2}{2}$. The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$,respectively. Then:
$(A)$ $a_x=1 \text{ m/s}^2$ implies that when the particle is at the origin,$a_y=1 \text{ m/s}^2$
$(B)$ $a_x=0$ implies $a_y=1 \text{ m/s}^2$ at all times
$(C)$ at $t=0$,the particle's velocity points in the $x$-direction
$(D)$ $a_x=0$ implies that at $t=1 \text{ s}$,the angle between the particle's velocity and the $x$-axis is $45^{\circ}$