$A$ variable line passing through $(l, m)$ intersects the coordinate axes at the points $A$ and $B$. If the line drawn parallel to $Y$-axis through $A$ and parallel to $X$-axis through $B$ meet at $P$,then the locus of $P$ 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 $45^{\circ}$

If a straight line drawn through the point of intersection of the lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$ meets the coordinate axes at the points $P$ and $Q$, then the locus of the midpoint of $PQ$ is:

$A$ line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

$A$ point $P$ moves inside a square of area $4$ square units such that it is nearer to the point of intersection of its diagonals than any vertex. The area of the region traced by $P$ is

$A$ straight rod of length $4$ units slides such that its ends $A$ and $B$ always lie on the $X$ and $Y$-axes respectively. Then,the locus of the centroid of $\triangle OAB$ is