Prove the locus of $M$ is a conic passing though $I$ and $J$.

Question

Problem: In the perspective plane $\mathbb{P}^2$, given a conic $(S)$ and a line $d$ having no intersection with $(S)$. Fixed two points $I$ and $J$ on $d$. Take a variety point $M$ such that the tangents from $M$ to $(S)$ intersect $d$ at two points $R$ and $S$ such that $(IJRS)=-1$. Prove the locus of $M$ is a conic passing through $I$ and $J$.

Here is what I've discovered so far.

1. Let the tangents from $I$ and $J$ touch $(S)$ at $I_1,I_2$ and $J_1,J_2$ respectively. Then the locus of $M$ is the conic passing through $I,I_1,I_2,J,J_1,J_2$. I also proved those 6 points lie on a conic (let's call it $(S')$).

2. Take any point on $(S')$, call $A$. Then the intersection $A_1, A_2$ of two tangents from $A$ to $(S)$ and $d$ satisfy $(IJ,A_1A_2)=-1$. Now I'm stucking at proving this claim.

2'. Something that might be useful for proving 2: Let $P$ the polar of $d$ wrt $(S)$. Then $PI$ and $PJ$ tangent to $(S')$.

Can I have some hints for the next steps? Any ideas would be appreciated. Thanks for reading.

Answer 1

The other answer here uses a proof by Milne that claims "range (c)=range(d)". What does that mean?

On page 29 the notations "range (a)" and "pencil P(a)" are defined. For our purposes, a "range" is any collection of collinear points and a "pencil" is any collection of concurrent lines. Often ranges and pencils contain just four elements so that we can take cross ratios, but in general they contain an arbitrary number of elements.

For shortness we shall often denote a range (abcde ...) by (a), and a pencil P{abcde ...) by P(a)

(Chapter IV is worth a read, or at least a skim)

Unfortunately, I cannot find any explicit definition by Milne of what is meant by "range(a)=range(b)" or "pencil P(a)= pencil P'(a')". From context is seems to be that "range(a)=range(b)" means the two are homographic. In fact, on page 187 pencils and ranges are said to be equal (homographic) as in:

the pencil $\rho(m)$ = the range of poles $(m)$ = the pencil of polars $P(\mu)$

Does this help? Getting back to your question, I interpret

And the range (c) = the range (d), therefore the pencil A(P) = the pencil B(P),

as

"Because range(c) is homographic to range(d) the pencil A(P) is homographic to B(P)"

In all of the above, ranges and pencils are held to be homographic because they preserve cross ratios. In particular, you should verify that the conjugacy $c \to d$ preserves cross ratio and thus is homographic.

Answer 2

This locus is discussed in Milne, Cross Ratio Geometry, Article 185.

There, $P,A,B$ correspond to OP's $M,I,J$ and the condition that the lines $PA,PB$ are conjugate is equivalent to OP's $R,S$ being harmonic with $I,J$.

The proof relies on the locus conic being defined by the intersections of two homographic pencils centered at $A$ and $B$.

The excerpt is below. But the full reference is given above if background and definitions need to be consulted.

I've added a discussion of the claim "range ($c$) = range ($d$)" in a separate answer.