Two points $A(-a, 0)$ and $B(a, 0)$ are given. If $C$ is a variable point lying on one side of the line $AB$ such that $\angle CAB - \angle CBA = \alpha$,where $\alpha$ is a positive constant,then the locus of the point $C$ is

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines passing through $A$ and $B$ and parallel to the coordinate axes is:

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

If the sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$,show that $P$ must move on a line.

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 $45^{\circ}$

$ABC$ is a variable triangle such that $A$ is $(1, 2)$,and $B$ and $C$ lie on the line $y = x + \lambda$ (where $\lambda$ is a variable). The locus of the orthocenter of triangle $ABC$ is: