If a line $AB$ of length $r$ moves so that $A$ and $B$ always lie respectively on the $X$-axis and the line $y=6x$,then the locus of the mid-point of $AB$ is:

$A$ ray of light coming from the point $(1, 2)$ is reflected at a point $A$ on the $x$-axis and then passes through the point $(5, 3)$. The coordinates of the point $A$ are

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 $45^{\circ}$

$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

Every point $(x, y)$ on the curve $3x + 2y - 3xy = 0$ is the centroid of a triangle formed by the coordinate axes and a line $(L)$ intersecting both the coordinate axes. Then all such lines $(L)$

$A$ light ray emerging from a point source at $A(2,3)$ is reflected on the $Y$-axis at point $B$ and passes through point $C(5,10)$. Then the coordinates of $B$ are: