Suppose a point $P$ moves so that $BP^2 - AP^2 = 121$,where $A$ and $B$ are $(2, 5)$ and $(5, 11)$ respectively. Then the locus of $P$ is a straight line,whose slope is

The line $AB$ cuts off equal intercepts $2a$ from the axes. From any point $P$ on the line $AB$, perpendiculars $PR$ and $PS$ are drawn to the axes. The locus of the mid-point of $RS$ is

Given points $A(6,0)$,$B(0,4)$,and $O$ as the origin,find the locus of a point $P(x, y)$ such that the area of $\triangle POB$ is $2$ times the area of $\triangle POA$.

The locus of a point $P(x, y)$ which moves in such a way that the segment $OP$,where $O$ is the origin $(0, 0)$,has a slope of $\sqrt{3}$ is:

If $ABC$ is an isosceles triangle and the coordinates of the base points are $B(1, 3)$ and $C(-2, 7)$, then the coordinates of $A$ can be:

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is a constant. Then,the locus of the foot of the perpendicular from the origin on the given line is -