A rod slides with its ends on two coordinate axes. A point $P$ divides the rod in the ratio $1:2$. Find the locus of $P$.

Let $A$ be the fixed point $(0,4)$ and $B$ be a moving point $(2t, 0)$. Let $M$ be the mid-point of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the mid-point $P$ of $MR$ is
(A) $y + x ^ { 2 } = 2$
(B) $x ^ { 2 } + ( y - 2 ) ^ { 2 } = 1 / 4$
(C) $( y - 2 ) ^ { 2 } - x ^ { 2 } = 1 / 4$
(D) none of the above

Let $L$ be the point $(t, 2)$ and $M$ be a point on the $y$-axis such that $LM$ has slope $-t$. Then the locus of the midpoint of $LM$, as $t$ varies over all real values, is
(A) $y = 2 + 2x^2$
(B) $y = 1 + x^2$
(C) $y = 2 - 2x^2$
(D) $y = 1 - x^2$

Let $A$ be the fixed point $(0,4)$ and $B$ be a moving point $(2t, 0)$. Let $M$ be the mid-point of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the mid-point $P$ of $MR$ is
(A) $y + x ^ { 2 } = 2$
(B) $x ^ { 2 } + ( y - 2 ) ^ { 2 } = 1 / 4$
(C) $( y - 2 ) ^ { 2 } - x ^ { 2 } = 1 / 4$
(D) none of the above

Let $L$ be the point $(t, 2)$ and $M$ be a point on the $y$-axis such that $LM$ has slope $-t$. Then the locus of the midpoint of $LM$, as $t$ varies over all real values, is
(A) $y = 2 + 2x^2$
(B) $y = 1 + x^2$
(C) $y = 2 - 2x^2$
(D) $y = 1 - x^2$

Let $A$ be the fixed point $(0,4)$ and $B$ be a moving point $(2t, 0)$. Let $M$ be the mid-point of $A B$ and let the perpendicular bisector of $A B$ meet the $y$-axis at $R$. The locus of the mid-point $P$ of $M R$ is
(A) $y + x ^ { 2 } = 2$
(B) $x ^ { 2 } + ( y - 2 ) ^ { 2 } = 1 / 4$
(C) $( y - 2 ) ^ { 2 } - x ^ { 2 } = 1 / 4$
(D) none of the above

Let $L$ be the point $(t, 2)$ and $M$ be a point on the $y$-axis such that $L M$ has slope $-t$. Then the locus of the midpoint of $L M$, as $t$ varies over all real values, is
(A) $y = 2 + 2 x ^ { 2 }$
(B) $y = 1 + x ^ { 2 }$
(C) $y = 2 - 2 x ^ { 2 }$
(D) $y = 1 - x ^ { 2 }$

A moving line intersects the lines $x + y = 0$ and $x - y = 0$ at the points $A$ and $B$ such that the area of the triangle with vertices $(0,0)$, $A$ and $B$ has a constant area $C$. The locus of the midpoint of $AB$ is given by the equation
(A) $\left(x^2 + y^2\right)^2 = C^2$
(B) $\left(x^2 - y^2\right)^2 = C^2$
(C) $(x + y)^2 = C^2$
(D) $(x - y)^2 = C^2$.

Let $A$ and $B$ be variable points on $x$-axis and $y$-axis respectively such that the line segment $AB$ is in the first quadrant and of a fixed length $2d$. Let $C$ be the mid-point of $AB$ and $P$ be a point such that
(a) $P$ and the origin are on the opposite sides of $AB$ and,
(b) $PC$ is a line segment of length $d$ which is perpendicular to $AB$.
Find the locus of $P$.

Consider the parabola $C: y^2 = 4x$ and the straight line $L: y = x + 2$. Let $P$ be a variable point on $L$. Draw the two tangents from $P$ to $C$ and let $Q_1$ and $Q_2$ denote the two points of contact on $C$. Let $Q$ be the mid-point of the line segment joining $Q_1$ and $Q_2$. Find the locus of $Q$ as $P$ moves along $L$.

The straight line $O A$ lies in the second quadrant of the $(x, y)$-plane and makes an angle $\theta$ with the negative half of the $x$-axis, where $0 < \theta < \frac { \pi } { 2 }$. The line segment $C D$ of length 1 slides on the $(x, y)$-plane in such a way that $C$ is always on $O A$ and $D$ on the positive side of the $x$-axis. The locus of the mid-point of $C D$ is
(A) $x ^ { 2 } + 4 x y \cot \theta + y ^ { 2 } \left( 1 + 4 \cot ^ { 2 } \theta \right) = \frac { 1 } { 4 }$.
(B) $x ^ { 2 } + y ^ { 2 } = \frac { 1 } { 4 } + \cot ^ { 2 } \theta$.
(C) $x ^ { 2 } + 4 x y \cot \theta + y ^ { 2 } = \frac { 1 } { 4 }$.
(D) $x ^ { 2 } + y ^ { 2 } \left( 1 + 4 \cot ^ { 2 } \theta \right) = \frac { 1 } { 4 }$.

The locus of the orthocentre of the triangle formed by the lines $$\begin{aligned} &(1+p)x-py+p(1+p)=0\\ &(1+q)x-qy+q(1+q)=0 \end{aligned}$$ and $y=0$, where $p\neq q$, is
(A) a hyperbola
(B) a parabola
(C) an ellipse
(D) a straight line

Let $( x , y )$ be any point on the parabola $y ^ { 2 } = 4 x$. Let $P$ be the point that divides the line segment from $( 0,0 )$ to $( x , y )$ in the ratio $1 : 3$. Then the locus of $P$ is
(A) $x ^ { 2 } = y$
(B) $y ^ { 2 } = 2 x$
(C) $y ^ { 2 } = x$
(D) $x ^ { 2 } = 2 y$

Let $\mathbb { R } ^ { 3 }$ denote the three-dimensional space. Take two points $P = ( 1,2,3 )$ and $Q = ( 4,2,7 )$. Let $\operatorname { dist } ( X , Y )$ denote the distance between two points $X$ and $Y$ in $\mathbb { R } ^ { 3 }$. Let
$$\begin{gathered} S = \left\{ X \in \mathbb { R } ^ { 3 } : ( \operatorname { dist } ( X , P ) ) ^ { 2 } - ( \operatorname { dist } ( X , Q ) ) ^ { 2 } = 50 \right\} \text { and } \\ T = \left\{ Y \in \mathbb { R } ^ { 3 } : ( \operatorname { dist } ( Y , Q ) ) ^ { 2 } - ( \operatorname { dist } ( Y , P ) ) ^ { 2 } = 50 \right\} \end{gathered}$$
Then which of the following statements is (are) TRUE?
(A) There is a triangle whose area is 1 and all of whose vertices are from $S$.
(B) There are two distinct points $L$ and $M$ in $T$ such that each point on the line segment $L M$ is also in $T$.
(C) There are infinitely many rectangles of perimeter 48, two of whose vertices are from $S$ and the other two vertices are from $T$.
(D) There is a square of perimeter 48, two of whose vertices are from $S$ and the other two vertices are from $T$.

Locus of the image of the point $( 2,3 )$ in the line $( 2 x - 3 y + 4 ) + k ( x - 2 y + 3 ) = 0 , k \in \mathbb{R}$, is a
(1) Circle of radius $\sqrt { 3 }$
(2) Straight line parallel to $x$-axis.
(3) Straight line parallel to $y$-axis.
(4) Circle of radius $\sqrt { 2 }$

If a variable line drawn through the intersection of the lines $\frac { x } { 3 } + \frac { y } { 4 } = 1$ and $\frac { x } { 4 } + \frac { y } { 3 } = 1$, meets the coordinate axes at $A$ and $B$, $( A \neq B )$, then the locus of the midpoint of $AB$ is:
(1) $7 x y = 6 ( x + y )$
(2) $4 ( x + y ) ^ { 2 } - 28 ( x + y ) + 49 = 0$
(3) $6 x y = 7 ( x + y )$
(4) $14 ( x + y ) ^ { 2 } - 97 ( x + y ) + 168 = 0$

A straight line through a fixed point $( 2,3 )$ intersects the coordinate axes at distinct points $P$ and $Q$. If $O$ is the origin and the rectangle $O P R Q$ is completed, then the locus of $R$ is:
(1) $3 x + 2 y = 6 x y$
(2) $3 x + 2 y = 6$
(3) $2 x + 3 y = x y$
(4) $3 x + 2 y = x y$

A point $P$ moves on the line $2 x - 3 y + 4 = 0$. If $Q ( 1,4 )$ and $R ( 3 , - 2 )$ are fixed points, then the locus of the centroid of $\triangle P Q R$ is a line:
(1) with slope $\frac { 2 } { 3 }$
(2) with slope $\frac { 3 } { 2 }$
(3) parallel to $y$-axis
(4) parallel to $x$-axis

The locus of the mid-points of the perpendiculars drawn from points on the line $x = 2 y$, to the line $x = y$, is.
(1) $2 x - 3 y = 0$
(2) $5 x - 7 y = 0$
(3) $3 x - 2 y = 0$
(4) $7 x - 5 y = 0$

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 co-ordinates axes, then $\alpha$ is equal to
(1) - 7
(2) $- 7 \sqrt { 3 }$
(3) $7 \sqrt { 3 }$
(4) 7

Let $A ( - 1,1 )$ and $B ( 2,3 )$ be two points and $P$ be a variable point above the line $A B$ such that the area of $\triangle \mathrm { PAB }$ is 10 . If the locus of P is $\mathrm { a } x + \mathrm { b } y = 15$, then $5 \mathrm { a } + 2 \mathrm {~b}$ is :
(1) 6
(2) $- \frac { 6 } { 5 }$
(3) 4
(4) $- \frac { 12 } { 5 }$

Let the points $\left( \frac{11}{2}, \alpha \right)$ lie on or inside the triangle with sides $x + y = 11$, $x + 2y = 16$ and $2x + 3y = 29$. Then the product of the smallest and the largest values of $\alpha$ is equal to:
(1) 44
(2) 22
(3) 33
(4) 55