If a variable line drawn through the intersection of the lines $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$ meets the coordinate axes at $A$ and $B$ $(A \neq B)$,then the locus of the midpoint of $AB$ 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 $45^{\circ}$

$A$ ray of light passing through the point $(1, 2)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $(5, 3)$. Find the coordinates of $A$.

If $P_1, P_2, P_3, \ldots, P_n$ are $n$ points on the line $y=x$ all lying in the first quadrant,such that $OP_n = n(OP_{n-1})$ ($O$ is origin),$OP_1 = 1$ and $P_n = (2520 \sqrt{2}, 2520 \sqrt{2})$,then $n=$

$A$ line segment joining a point $A$ on $x$-axis to a point $B$ on $y$-axis is such that $AB=15$. If $P$ is a point on $AB$ such that $\frac{AP}{PB}=\frac{2}{3}$,then the locus of $P$ is:

$A \equiv (\cos \theta, \sin \theta)$ and $B \equiv (\sin \theta, -\cos \theta)$ are two points. The locus of the centroid of $\triangle OAB$,where $O$ is the origin,is