C1.
a. Determine all points $P(x, y)$ such that $(0,0),(1,1),(1,0)$ and $P$ are vertices of a parallelogram.
b. Two parallel lines intersect the (horizontal) parabola $x=y^{2}$ at four distinct points: $(0,0)$, $(1,1),(9,3)$ and $Q$. Determine all possible coordinates of the point $Q$.
c. Two parallel lines intersect the parabola $x=y^{2}$ at four distinct points: $(0,0),(1,1),\left(a^{2}, a\right)$ and $V$. Here $a \neq 0, \pm 1$ is a real number. Determine all possible coordinates of the point $V$. The answer should be expresed in term of $a$.
Solution:
a. There are three possibilities for $P:(0,1),(0,-1),(2,1)$.
b. There are three possible pairs of parallel lines - the line through $(0,0)$ can go through $(1,1)$, or $(9,3)$ or $Q$.
- The line through $(0,0)$ and $(1,1)$ is parallel to $y=x-6$ through $(9,3)$. The solutions to $y+6=y^{2}$ give the $y$ coordinates of the intersection points as $x=y+6$ and $x=y^{2}$. Hence at $y=3$ or $y=-2$, so the $y$ coordinate of $Q$ here is -2 and then $x=4$.
- The line through $(0,0)$ and $(9,3)$ is parallel to $y=\frac{x+2}{3}$ through $(1,1)$. The solutions to $3 y-2=y^{2}$, are $y=1$ or $y=2$, so the $y$ coordinate of $Q$ here is 2 and $x=4$.
- The line through $(1,1)$ and $(9,3)$ is parallel to $y=\frac{x}{4}$ through $(0,0)$. The non-zero solution to $4 y=y^{2}$ is $y=4$, so the $y$ coordinate of $Q$ here is 4 and $x=16$.
Thus, $Q$ could be at $(4,-2),(4,2),(16,4)$.
c. There are three possible pairs of parallel lines - the line through $(0,0)$ can go through $(1,1)$, or $\left(a^{2}, a\right)$ or $V$.
- The line through $(0,0)$ and $(1,1)$ is parallel to $y=x+a-a^{2}$ through $\left(a^{2}, a\right)$. The solutions to $y+a(a-1)=y^{2}$ give the $y$ coordinates of the intersection points, hence at $y=a$ or $y=1-a$, so the $y$ coordinate of $V$ here is $1-a$ and then $x=(1-a)^{2}$.
- The line through $(0,0)$ and $\left(a^{2}, a\right)$ is parallel to $y=\frac{x+a-1}{a}$ through $(1,1)$. The solutions to $a y+1-a=y^{2}$, are $y=1$ or $y=a-1$, so the $y$ coordinate of $V$ here is $a-1$ and $x=(a-1)^{2}$.
- The line through $(1,1)$ and $\left(a^{2}, a\right)$ is parallel to $y=\frac{x}{a+1}$ through $(0,0)$. The non-zero solution to $(a+1) y=y^{2}$ is $y=a+1$, so the $y$ coordinate of $V$ here is $a+1$ and $x=(a+1)^{2}$.
Thus, $V$ could be at $\left((a-1)^{2}, \pm(a-1)\right.$ and $\left.\left((a+1)^{2}, a+1\right)\right)$.
A variation of the solution for $C 1(c)$ : There are three possible pairs of parallel lines - the line through $(0,0)$ can go through $(1,1)$, or $\left(a^{2}, a\right)$ or $V$.
- The line through $(0,0)$ and $(1,1)$ has slope 1. A line through $\left(a^{2}, a\right)$ parallel to it has the equation $\frac{y-a}{x-a^{2}}=1$. Substitute $x=y^{2}$ in order to find the points of intersection of the line with the parabola. We have $\frac{y-a}{y^{2}-a^{2}}=1$ or $\frac{y-a}{(y-a)(y+a)}=1$. To exclude the original point we assume $y \neq a$. Thus, we can cancel $y-a$ and get $y+a=1$, so $y=1-a$. Then $x=(1-a)^{2}$.
- The line through $(0,0)$ and $\left(a^{2}, a\right)$ has slope $\frac{1}{a}$. A line through $(1,1)$ parallel to it has the equation $\frac{y-1}{x-1}=\frac{1}{a}$. Substitute $x=y^{2}$ in order to find the points of intersection of the line with the parabola. We have $\frac{y-1}{y^{2}-1}=\frac{1}{a}$ or $\frac{y-1}{(y-1)(y+1)}=\frac{1}{a}$. To exclude the original point we assume $y \neq 1$. Thus, we can cancel $y-1$ and get $y+1=a$, so $y=a-1$. Then $x=(a-1)^{2}$.
- The line through $(1,1)$ and $\left(a^{2}, a\right), a \neq \pm 1$ has slope $\frac{a-1}{a^{2}-1}=\frac{1}{a+1}$. A line through $(0,0)$ parallel to it has the equation $\frac{y}{x}=\frac{1}{a+1}$. Substitute $x=y^{2}$ in order to find the points of intersection of the line with the parabola. We have $\frac{y}{y^{2}}=\frac{1}{a+1}$. To exclude the original point we assume $y \neq 0$. Thus, we can cancel $y$ and get $y=a+1$. Then $x=(a+1)^{2}$.
Thus, $V$ could be at $\left((a-1)^{2}, \pm(a-1)\right.$ and $\left((a+1)^{2}, a+1\right)$. Finally, we should ensure that none of the points $(0,0),(1,1),\left(a^{2}, a\right)$ and $V$ coincide. The problem poses the restriction $a \neq 0$, $\pm 1$. In addition, we find that for $a=\frac{1}{2},\left((a-1)^{2},-(a-1)\right)=\left(a^{2}, a\right)=\left(\frac{1}{4},-\frac{1}{2}\right)$, and for $a=2$, $\left((a-1)^{2},(a-1)\right)=(1,1)$. Therefore the answer is: for $a \neq 0, \pm 1, \frac{1}{2}, 2$ point $V$ could be at $\left((a-1)^{2}, \pm(a-1)\right)$ and $\left((a+1)^{2}, a+1\right)$; for $a=\frac{1}{2}$ point $V$ could be at $\left(\frac{9}{4}, \frac{3}{2}\right)$ or $\left(\frac{1}{4},-\frac{1}{2}\right)$; for $a=2$ point $V$ could be at $(1,-1)$ or $(9,3)$. We also note that for $a=-1$, point $V$ could be at $(4,-2)$ or $(4,2)$.