I need to prove the following theorem:
Theorem Only one conic can be drawn having any two given parallel chords and its centre is at any point on the line bisecting the chords.
I tried to prove it using Analytic Geometry, but I failled. I supposed that the point $P_1(x_1,y_1)$ is in the first chord, that $(a,b)$ is a fixed direction and that the $x$-axis is the axis of the conic. In this way, the center is $C(c,0)$.
I found the point $V_1(x_1-\frac{a}{b}y_1,0)$ as the intersection between the chord and the axis. More than that, $Q_1(x_1-2\frac{a}{b}y_1,-y_1)$ is another intersection between the conic and the chord.
The same procedures can be done for the second chord, getting the points $P_2(x_2,y_2)$, $V_2(x_2-\frac{a}{b}y_2,0)$ (intersection between the chord and the axis) and $Q_2(x_2-2\frac{a}{b}y_2,-y_2)$ (intersection between the conic and the chord).
Thanks for any help.
Cleto
A well-known property of conics states that the midpoints of parallel chords lie on a line passing through the center.
Hence, given two parallel chords $AB$, $CD$ and the line passing through their midpoints, once you choose the center $O$ on that line the conic is fixed: if $A'$ is the reflection of $A$ about $O$, then there is a unique conic through $ABCDA'$ and that conic has $O$ as its center.
EDIT.
As noted by Blue in a comment, the theorem is false in the special case $AB\cong CD$, if center $O$ is chosen at the center of parallelogram $ABDC$: in that case any conic through $ABCD$ satisfies the given conditions. If center $O$ is not the center of parallelogram $ABDC$, the conic is unique but degenerates into a pair of parallel lines.