$A$ line cuts the $X$-axis at $A(5,0)$ and the $Y$-axis at $B(0,-3)$. $A$ variable line $PQ$ is drawn perpendicular to $AB$ cutting the $X$-axis at $P$ and the $Y$-axis at $Q$. If $AQ$ and $BP$ meet at $R$, then the locus of $R$ 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 $45^{\circ}$

$A$ light ray emerging from the point source placed at $P(1, 3)$ is reflected at a point $Q$ on the $x$-axis. If the reflected ray passes through the point $R(6, 7)$,then the abscissa of $Q$ is

The locus of a point which moves so that it is always equidistant from the points $A(a, 0)$ and $B(-a, 0)$ is

If the line segment joining the points $(1,0)$ and $(0,1)$ subtends an angle of $45^{\circ}$ at a variable point $P$,then the equation of the locus of $P$ is

If the equation of the locus of a point equidistant from the points $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,then the value of $c$ is