Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$,then the equation of the locus of the midpoint $P$ of $BC$ is

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7, \theta \in \left(0, \frac{\pi}{2}\right)$ between the coordinate axes,then $\alpha$ is equal to

The coordinates of the points $A$ and $B$ are $(a, 0)$ and $(-a, 0)$ respectively. If a point $P$ moves such that $PA^2 - PB^2 = 2k^2$,where $k$ is a constant,then the equation to the locus of the point $P$ is:

If $P = (1, 0)$,$Q = (-1, 0)$,and $R = (2, 0)$ are three given points,then the locus of the points $S$ satisfying the relation $SQ^2 + SR^2 = 2 SP^2$ is :

Let $BC$ be a fixed line segment in the plane. The locus of a point $A$ such that the $\triangle ABC$ is isosceles,is (with finitely many possible exceptional points)

$A$ 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