Why is each chord of a conic having a unique midpoint corresponding to it? (Excluding centre of symmetries)

Question

If you have a conic $C(x,y)$ then by homogenization, we find the tangent line at the point $(x,y)$ to be $L(x,y,X,Y,Z=1)=\frac{\partial C}{\partial x} X + \frac{\partial C}{\partial y} Y + \frac{\partial C}{\partial Z}=0$, now let $(\alpha,\beta)$ and $(\alpha',\beta')$ be two distinct point on the conic and $(l,n)$ their midpoint, then it turns out that:

$$C(l,n)= L(l,n,X,Y,Z=1)$$

To be the equation of chord of point $(\alpha,\beta)$ and $(\alpha',\beta')$ but is there any way to prove that any two points on a conic have a unique midpoint?

I find the uniqueness to be quite a crazy result, since it associates each chord on the conic to a point and hence want a bit more insight on it.

Observation:

The equation fails if you take a point on center of symmetery of the conic eg: center of circle, there are many points mapping to the same mid point.. for sake of original question, leave this. But, I am also curious to how this equation breaks down from algebraic point of view as well.

Interesting corollaries:

For a unit circle parameterized as $(\cos t,\sin t)$ , it turns out that if we have four points on the circle $\alpha,\beta,\gamma, \delta$, and we consider MP point corresponding to $\alpha$ and $\beta$ and also point corresponding to $\gamma$ and $\delta$, then the equation:

$$ \cos \alpha + \cos \beta = \cos \gamma + \cos \delta$$

Has no solution unless the angles other than the case where angles are diametrically opposite i.e: different by $\pm \pi$. Similar result for $\sin$.

The equation of tangent : If we let $(l,n)$ be a point on the conic, we recover equation of tangent of conic at a point.

Please note, it is the partial derivatives which are function of $(x,y)$ in the equation of line.

Examples of usage : 2

Homogenization can be found in an old calculus book by Joseph Edwards

Are you talking about points of contact from an external point or any two pairs of points on the conic having unique midpoint? — Math Lover, Aug 07 '21 at 18:27
The equation relates to the chord line through mid point, so any pair of point-> a unique mid point( apprently)-> a unique chord @MathLover — tryst with freedom, Aug 07 '21 at 18:30
yes now the question seems complete (after you added the observation) — Math Lover, Aug 07 '21 at 18:46
Note that at the center of the circle, the gradient of $C$ vanishes. — Ted Shifrin, Aug 07 '21 at 18:51
This property follows from a well-known fact: all parallel chords of a conic have their midpoints aligned along a diameter passing through the center (or parallel to the axis in the case of a parabola). Tangents at the endpoints of the diameter are also parallel to the chords. Search "conjugate diameters of a conic" to find a variety of proofs. — Intelligenti pauca, Aug 07 '21 at 19:24

Intelligenti pauca · Answer 1 · 2021-08-08T08:36:01.250

1

If point $P$ is within a conic section (but different from its centre) then there is a unique chord having $P$ as midpoint.

Proof. Join $P$ with the centre of the conic (or draw the line through $P$ parallel to the axis, in the case of a parabola) and produce that line (if needed) to meet the conic at $A$. By the properties of conjugate diameters, the chord through $P$ parallel to the tangent at $A$ is the unique chord having midpoint $P$.

For a hyperbola, this also works if $P$ is outside the hyperbola (but not on an asymptote), provided we also consider the conjugate hyperbola.

edited Aug 08 '21 at 08:36

answered Aug 07 '21 at 19:45

Intelligenti pauca

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And the chord eq is given by $T=S_1$ – DatBoi Aug 07 '21 at 20:01

Why is each chord of a conic having a unique midpoint corresponding to it? (Excluding centre of symmetries)

Observation:

1 Answers1