Given collinear points $A$, $B$, $C$, a variable line through $C$ meets a conic at $P$ and $Q$. Why does $AP\cap BQ$ trace another conic?

Question

At first, there are a fixed conic curve $\Gamma$ and three fixed collinear points $A,B,C$. Draw a moving straight line $l$ through point C, intersecting $\Gamma$ at two points $P,Q$. Find the two intersections $S=AP\cap BQ$ and $T=AQ\cap BP$. As line $l$ move, $S$ and $T$ generate a locus $\Delta$. Due to symmetry, $S$ and $T$ must have a common locus.

I found that $\Delta$ will be a new conic curve. So the question is: why this is true?

Use knowledge harmonic point sequences, the intersection of $ST$ and $AB$ is a fixed point $D$ that form a harmonic point sequences $[A,B;C,D]$ together with points $A,B,C$.

Suppose that $C$ is outside of $\Gamma$. Make two tangent lines $CM,CN$ from $C$ to $\Gamma$. Then the line $MN$ is the polar of $C$ in $\Gamma$. Let $E$ be the intersection of $MN$ and $l$. Through the image we can guess these two are right:

1. $E$ is also the intersection of $MN$ and $ST$.
2. $DM,DN$ also tangent to $\Delta$.

The proof of the first guess is not too hard. Let $E'$ be the intersection of $ST$ and $l$, then all that need to do is to prove $E=E'$. According to the property of pole $C$ and polar $MN$, $[C,E;P,Q]$ is a harmonic point sequences; according to the property of complete quadrangle $[A,B;P,Q;S,T]$, $[C,E';P,Q]$ is also a harmonic point sequences; these two facts indecate $E=E'$.

The second guess suggest that $[C,\Gamma,P,Q]$ and $[D,\Delta,S,T]$ may be symmetric. This inspired me to turn to find that what's the relation between $\Gamma$ and $C,M,N$. Or, if three points $C,M,N$ have been given, what conditions should a point $P$ meets in order to the locus of $P$ is a conic curve tangenting $CM,CN$ at $M,N$. If this is finded, we can transfer the condition from $P$ or $Q$ to $S$ or $T$, then $S$ or $T$ must meet the same condition and the locus of $S,T$ must be a conic curve tangenting $DM,DN$ at $M,N$.

The ratio $\dfrac{\overline{ME}}{\overline{EN}}$ uniquely determines the position of $E$ on $MN$; the position of $E$ determine the line $l$; and the other ratio $\dfrac{\overline{CP}}{\overline{PE}}$ also uniquely determines the position of $P$ on $l$. So these two ratio must have a relation when $P$ moving on a fixed conic curve $\Gamma$. By attempting on some special curve, I find this theorem:

Theorem: Let $E$ be a moving point on $MN$, and $P$ a point on $CE$, then the locus of $P$ is a conic curve if and only if this equation holds: $$ \left(\frac{\overline{CP}}{\overline{PE}}\right)^2\frac{\overline{ME}\overline{EN}}{\overline{MN}^2}=\text{Const}$$

If the Theorem is true, by Menelaus's Theorem on triangle $\triangle CED$ and transversal line $PSA$ we know $\dfrac{\overline{CP}}{\overline{PE}}\times \dfrac{\overline{ES}}{\overline{SD}}\times \dfrac{\overline{DA}}{\overline{AC}}=-1$. So there is $\dfrac{\overline{DS}}{\overline{SE}}= -\dfrac{\overline{DA}}{\overline{AC}} \times\dfrac{\overline{CP}}{\overline{PE}}$ in which $-\dfrac{\overline{DA}}{\overline{AC}}$ is contant.Then $$ \left(\frac{\overline{DS}}{\overline{SE}}\right)^2 \frac{\overline{ME}\overline{EN}}{\overline{MN}^2}= \left(\frac{\overline{DA}}{\overline{AC}}\right)^2 \left(\frac{\overline{CP}}{\overline{PE}}\right)^2 \frac{\overline{ME}\overline{EN}}{\overline{MN}^2} =\text{Const}$$ and give what I want.

Now I have these questions:

1. How to prove the above theorem?
2. How to explain the above theorem?
3. How to prove the same problem when $C$ is inside of $\Gamma$?, excluding use the space $\mathbb{CP}^2$. Is there another way to prove the problem
4. Do I find this fact that collinear $A,B,C$ transform $\Gamma$ to a new conic curve $\Delta$ first? Is it been found by someone before?

Answer 1

Now I have a proof of the question based on projective geometry.

Using homogeneous coordinates, and we argee on that the symbol of point $X$ or line $l$ also represent its coordinates.

We choose appropriate multiples of coordinates of $C,D$ that satisfies $A=C+D$, then because $[A,B;C,D]$ is a harmonic sequence, the coordinate of B can be represent as $B=C-D$. Treating $[AB]$ as a degenerated conic envelope, its matrix is $[AB]=AB^T+BA^T=CC^T-DD^T$.

We take an arbitrary line $m$ passing $C$, so that they satisfy $mC=0$.

We represent the matrix of conic curve $\Gamma$ as also $\Gamma$, then the matrix of conic envelope $\Gamma$ should be $\Gamma^{-1}$.

Lemma 1: If line $m$ and conic curve $\Gamma$ intersect at point $P,Q$, by treating $[PQ]$ as a degenerated conic envelope, its matrix will be $[PQ]=\Gamma^{-1}m^Tm\Gamma^{-1}-(m\Gamma^{-1}m^T)\Gamma^{-1}$.
Proof: A line $l$ belonging to $[PQ]$ is equivalent to the intersection $S$ of $m$ and $l$ belonging to conic curve $\Gamma$. And $S$ belonging to conic curve $\Gamma$ is equivalent to that there is only one line $t$ passing $S$, which can be represented as $t=m+zl,\ z\in\mathbb C$, belonging to conic envelope $\Gamma$, which can be represented as $t\Gamma^{-1}t^T=0$.
So the uniqueness means that the equation $$(m+zl)\Gamma^{-1}(m+zl)^T=(l\Gamma^{-1} l^T)z^2 + 2(m\Gamma^{-1}l)z + m\Gamma^{-1}m^T = 0$$ has only one solution of $z$, and the discriminant equation is $$\Delta=4(m\Gamma^{-1}l)^2-4(l\Gamma^{-1} l^T)(m\Gamma^{-1}m^T)=0.$$ So l belonging to $[PQ]$ is equivalent to $$l^T(\Gamma^{-1}mm^T\Gamma^{-1})l-l^T[(m\Gamma^{-1}m^T)\Gamma^{-1}]l=0$$ and the matrix of $[PQ]$ is $$[PQ]=\Gamma^{-1}m^Tm\Gamma^{-1}-(m\Gamma^{-1}m^T)\Gamma^{-1}.$$

Lemma 2: If there is a complete quadrangle $[A,B;P,Q;S,T]$, treating $[AB],[PQ],[ST]$ as degenerated conic envelopes, their matrices $[AB],[PQ],[ST]$ satisfy that $$\exists\lambda,\mu,\nu,\ \ s.t.\ \lambda[AB]+\mu[PQ]+\nu[ST]=0$$ Proof: As long as we realize that the four lines $\overline{APS}, \overline{AQT}, \overline{BPT}, \overline{BQS}$ belong to all three envelopes $[AB],[PQ],[ST]$, we should know that $[AB],[PQ],[ST]$ belong to a family of conic envelopes and therefore it holds that $\lambda[AB] + \mu[PQ] + \nu[ST] = 0$ with certain numbers $\lambda,\mu,\nu$.

Because $A,B$ is not on $\Gamma$, we know that $[AB]$ and $[PQ]$ cannot be the same. Then combining those two lemma, we know that $$\exists\lambda,\mu, [ST]=\lambda[AB]+\mu[PQ]=\lambda(CC^T-DD^T) + \mu\left[\Gamma^{-1}m^Tm\Gamma^{-1}-(m\Gamma^{-1}m^T)\Gamma^{-1}\right]$$

And we have another limitation that $[ST]$ is degenerated, which means there exists a line $n$ satisfied $n[ST]=0$. If fact, $n$ is just the line passing $S$ and $T$. Based on the geometric image, we know that $n$ pass $D$ and $E$, in which $E$ is the intersection of $m$ and $p$, the polar line of $C$ with repect to $\Gamma$, with coordinate $p=C^T\Gamma$. Because $n$ pass $E$, it holds that $n=\alpha m+\beta p=\alpha m+\beta C^T\Gamma$. And because $n$ pass $D$, we know $0=nD=\alpha mD+\beta C^T\Gamma D$ and can set $\alpha=C^T\Gamma D,\beta=-mD,n=(C^T\Gamma D)m-(mD)C^T\Gamma$.

We calculate that (notice that $mC=0,nD=0,m[PQ]=0$) $$ n[ST]=-\lambda(mD)(C^T\Gamma C)C^T+\mu(mD)(m\Gamma^{-1}m^T)C^T$$ So if we set $\lambda=m\Gamma^{-1}m^T,\mu=C^T\Gamma C$, we will get $n[ST]=0$ and then $[ST]$ is really a degenerated conic envelope with the matrix: $$[ST]=(m\Gamma^{-1}m^T)(CC^T-DD^T) + (C^T\Gamma C) \left[\Gamma^{-1}m^Tm\Gamma^{-1} - (m\Gamma^{-1}m^T)\Gamma^{-1}\right]\\ =(C^T\Gamma C)\Gamma^{-1}m^Tm\Gamma^{-1} - (m\Gamma^{-1}m^T)(C^T\Gamma C)\left[\Gamma^{-1} - \dfrac{CC^T - DD^T}{C^T\Gamma C}\right]$$

Then we point out the following lemma:

Lemma 3: With a nondegenerated conic curve $\Gamma$ and two points $S,T$, $S,T$ both on $\Gamma$ is equivalent to there exists a point $X$ satisfying $$\exists \alpha,\beta,\gamma,\ \ \alpha\Gamma^{-1}+\beta[ST]+\gamma XX^T=0$$ Proof On the one hand, if $S,T$ is both on $\Gamma$, let $p$ be the line $\overline{ST}$, then by Lemma 1 we have $$[ST]=\Gamma^{-1}p^Tp\Gamma^{-1}-(p\Gamma^{-1}p^T)\Gamma^{-1}.$$ Let $X=\Gamma^{-1}p^T$ and then we have $$ (p\Gamma^{-1}p^T)\Gamma^{-1} + [ST] - XX^T=0$$ and the conclusion holds.
On the other hand,if the conclusion holds, let $p=X^T\Gamma^{-1}$, and the matrix of intersections of $p$ and $\Gamma$, represented by $[R]$, satifies $$[R]=\Gamma^{-1}p^Tp\Gamma^{-1}-(p\Gamma^{-1}p^T)\Gamma^{-1} =XX^T-(X^T\Gamma X)\Gamma^{-1}$$ Because $\Gamma^{-1}$ is nondegenerated, and $XX^T$ has rank 1, if we think about $XX^T-\xi\Gamma^{-1}$ and demand it to be degenerated, we think with equation $\det(XX^T-\xi\Gamma^{-1})=0$ about $\xi$, then we will get three solutions of this cubic equation: $\xi_1=\xi_2=0$ giving $XX^T$ and $\xi_3=X^T\Gamma X$ giving $[R]$.
Now $-\beta[ST]=\alpha\Gamma^{-1}+\gamma XX^T$ is degenerated and different from $XX^T$, from that $[ST]$ should have rank 2, we know $[ST]$ is a multiple of $[R]$ and $S,T$ is just the intersections of $p$ and $\Gamma$.

Let $$\Delta^{-1}=\Gamma^{-1} - \frac{CC^T - DD^T}{C^T\Gamma C}$$ and $X=\Gamma^{-1}m^T$, then it holds that $$(m\Gamma^{-1}m^T)(C^T\Gamma C)\Delta^{-1} +[ST] - (C^T\Gamma C)XX^T =0 $$ So by Lemma 3, we know that $S,T$ is both on the fixed curve $\Delta$ and this has proven our problem.

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My English is not very good, and I am very sorry about that.

Answer 2

I get a straight line segment rather than a conic (in a special case). We can place the line $ABC$ at infinity, and assume that $C$ is the horizontal direction, whereas A and B are the directions of $e^{\pi i/3}$ and $e^{2\pi i/3}$. The intersection points P and Q are symmetric with respect to the $y$-axis, and therefore the intersection point $AP\cap BQ$ will always lie on the $y$-axis. This is of course a degenerate case.

In general, one can argue as follows. Use a projective transformation to send the line $ABC$ to the line at infinity. Then use an affine transformation to send the ellipse to the unit circle. Furthermore, by a rotation we can assume that $C$ corresponds to the pencil of horizontal lines (including the $x$-axis in the plane). Then the points $A$ and $B$ correspond to pencils of parallel lines of slopes $m_1, m_2\not=0$.

A horizontal line cuts the unit circle in (at most) two points $(s,\sqrt{1-s^2})$ and $(-s,\sqrt{1-s^2})$. We need to show that the intersection point $(x,y)$ of a pair of lines of slopes $m_1$ and $m_2$ passing through such a pair of points satisfies a quadratic equation. Then by definition it will be a conic section (at least in the nondegenerate case).

The equations of the two lines are $y=\sqrt{1-s^2} + m_1 (x+s)$ and $y=\sqrt{1-s^2} + m_2(x-s)$. It follows that $x=\frac{m_1+m_2}{m_2-m_1} s$ or $s=mx$ where $m=\frac{m_2-m_1}{m_1+m_2}$. From the equation of the first line we obtain $\big(y-m_1(x+mx)\big)^2= 1- (mx)^2$. This is a quadratic equation in $x$ and $y$, as required.

This argument works in the case when the conic and the line $ABC$ have no common points (either finite or infinite).

Your argument is longer but has the advantage of being almost synthetic ("almost" because you also use Menelaus theorem).

If the conic $K$ and the line $ABC$ have a single intersection point, one can apply a similar argument. We move the line $ABC$ to the line at infinity, and then move $C$ to the pencil of horizontal lines and the point $ABC \cap K$ to the pencil of vertical lines. We can then apply a translation to make sure that the axis of symmetry of the parabola $K$ is the $y$-axis and the apex of the parabola is at the origin. Afterwards apply a scaling to make sure the parabola is the standard parabola $y=x^2$. A horizontal line cuts the parabola at $(\pm s, s^2)$. The equations of the two lines of slopes $m_1$ and $m_2$ are then $y=s^2+m_1(x+s)$ and $y=s^2+m_2(x-s)$. We obtain a similar relation $s=mx$ (for the same $m$ as above). In this case also we obtain a quadratic relation between $x$ and $y$, of the form $y=(mx)^2 + m(x+mx)$, as required.