The locus of a variable point which forms a triangle of fixed area with two fixed points 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

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 an isosceles triangle. If the coordinates of the base are $(1, 3)$ and $(-2, 7)$,then the coordinates of vertex $A$ can be:

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,then its locus is:

If the equation of a curve remains unchanged by replacing $x$ with $y$ and $y$ with $x$,then the curve is