A point moves such that its distance from the point $(4,\,0)$is half that of its distance from the line $x = 16$. The locus of this point is

Consider the lines $L_1$ and $L_2$ defined by

$L _1: x \sqrt{2}+ y -1=0$ and $L _2: x \sqrt{2}- y +1=0$

For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.

Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.

($1$) The value of $\lambda^2$ is

($2$) The value of $D$ is

The locus of a point so that sum of its distance from two given perpendicular lines is equal to $2$ unit in first quadrant, is

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side a such that $a^{2}+11 a+3\left(m_{2}^{2}+m_{2}^{2}\right)=220$. If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$, where $\alpha \in\left(0, \frac{\pi}{2}\right)$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x +(\sin \alpha+\cos \alpha) y =10$, then  $72 \left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$ is equal to.

The sides of a rhombus $ABCD$ are parallel to the lines, $x – y + 2\, = 0$ and $7x – y + 3\, = 0$. If the diagonals of the rhombus intersect at $P( 1, 2)$ and the vertex $A$ ( different from the origin) is on the $y$ axis, then the ordinate of $A$ is