Prove that the line $XY$ goes through a fixed point where $X,Y$ are on fixed conic so that $\angle XPY = 90$ where $P$ is fixed on the conic.

Question

Say $\mathcal{C}$ is some conic and $P\in \mathcal{C}$ is fixed point on it. For each $X$ on $\mathcal{C}$ let $Y$ be such on $\mathcal{C}$ that $\angle XPY = 90^{\circ}$. Prove that the line $XY$ goes through a fixed point.

I can prove this with projective geometry:

Transformation $\Pi: PX\mapsto PY$ is projective from pencil of lines through $P$ to it self. Since $\Pi$ is induced by a rotation for $90^{\circ}$ around $P$ it is actually an involution. Now $\Pi$ induces new projective transformation $\pi$ from $\mathcal{C}$ to itself which is also an involution. Now there is a theorem which says that every involution on a conic is induced by some fixed point $F$ which lies on $XY$ and we are done.

Now, I'm interested in analytically and synthetically solution. Any help.

Edit: As you can notice on a picture, the fixed point $F$ is on a normal through $P$.

What fixed point does $XY$ go through? It's not $P$, since they form a right angle at $P$. — Paul, Apr 03 '20 at 17:13
The same question was already answered here: https://math.stackexchange.com/questions/1972621/prove-that-the-chord-of-the-ellipse-passes-through-a-fixed-point — Intelligenti pauca, Apr 03 '20 at 19:45
Not quite, this is general conic, while there is very specific ellipse. — nonuser, Apr 03 '20 at 19:47
Blue's answer there is for a generic ellipse, and it shouldn't be difficult to repeat the same reasoning for a parabola or hyperbola. — Intelligenti pauca, Apr 03 '20 at 19:57
No, it is not! Actually it is a point on a normal through $P$. A forgot to write this. But you can actually see that on a picture. — nonuser, Apr 13 '20 at 09:06
@Aqua: so we have basically to prove that the normal throught $P$ and the line $XY$ have an intersection if $\angle{XPY}=90°$? — Matteo, Apr 13 '20 at 10:04
@Aqua: so we have basically to prove that the normal throught $P$ and the line $XY$ have an intersection if $\angle{XPY}=90°$ and that intersection is fixed $\forall XY, P$ we consider? — Matteo, Apr 13 '20 at 10:09
@Aqua: So I guess my answer is not what you wanted and you are letting the bounty expire? — brainjam, Apr 13 '20 at 13:46
Not at all, I will do it in 1 hour before the and of it. Do not worry. @brainjam — nonuser, Apr 13 '20 at 13:47
Thanks! Btw, I prefer the projective proof you gave in OP to either the analytic or synthetic proof. It's short and -- because angles are not part of projective geometry -- surprising. — brainjam, Apr 13 '20 at 14:10
@brainjam Yes, but I can not use it in a highschool clasroom :( — nonuser, Apr 13 '20 at 14:49

brainjam · Answer 1 · 2020-04-06T06:02:26.767

4

There is an analytic proof in Salmon's Conic Sections, pg 165. He also shows that the fixed point is on the normal at $P$.

He also uses the analytic approach to generalize the proposition. If the angle is not a right angle, or if the angle is a right angle but the point is not on the conic, the chord will envelop a conic (see pg 267).

edited Apr 06 '20 at 06:02

answered Apr 06 '20 at 05:22

brainjam

9,172

brainjam · Answer 2 · 2020-04-07T01:10:40.060

There are synthetic proofs in 19th century texts. They are too long to reproduce here, but the texts are easily accessible with stable URLs, so I'll give a bare bones summary here and then some links and search terms so interested parties can find out more. The methods referred to here straddle projective and Euclidean geometry, but that's true for conics as well.

It's well known that a homography (aka projective transformation) takes lines to lines and conics to conics. Using the labels in OP, a homography takes $C$ to $C'$, $P$ to $P'$, $XY$ to $X'Y'$, and so on. It's well known that there are homographies that take $C$ to a circle. Let's assume that we can find a specific homography $T$ that maps angles at $P$ to equal angles at $P'$. I.e. $C'$ is a circle, and for all points $A,B$ $$\angle{APB}=\angle{A'P'B'}.$$

Then it's easy to see that chords $XY$ map to diameters $X'Y'$ of $C'$, which of course are concurrent at the center $O'$ of $C'$. It follows that the $XY$ all go through $O$ (the pre-image of $O'$).

The details of constructing $T$ and proving its properties are too lengthy to transpose here, but readable accounts can be found in Conic sections treated geometrically, by S.H. Haslam and J. Edwards, Chapter IX (where $T$ is called a focal projection) and Geometry of Conics by Charles Taylor, Sections 141-144, where $T$ is called a reversion.

The proposition of the OP is called Frégier's Theorem, and the fixed point is called Frégier's Point (in the context of determining an involution on a conic). The construction of $T$ goes back to Boscovich, 1754 (see also The Eccentric Circle of Boscovich).

Prove that the line $XY$ goes through a fixed point where $X,Y$ are on fixed conic so that $\angle XPY = 90$ where $P$ is fixed on the conic.

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