Can affine transformation make two intersective lines become parallel?

Question

The book Geometry of Conics, by Akopyan and Zaslavsky, states a problem as follows:

Probelem 9 A line intersects a hyperbola at points $P$ and $Q$ and its asymptotes at points $X$ and $Y$. Prove that the segments $PX$ and $QY$ are equal.

Besides, it gives a proof as follows:

Apply an affine transformation such that the line becomes parallel to one of the axes of the hyperbola. Then the desired equality would follow from symmetry. But since the segments in question lie on a line, their images under the affine transformation will also be equal.

It's well-known that the affine transformation preserves the parallelism, in another word, under such transformation, parallel lines are also parallel, and intersective lines are still intersective. How can an affine transformation make the line and the axis (not always parallel) become parallel?

Answer 1

The question is about

Problem 9 A line intersects a hyperbola at points $P$ and $Q$ and its asymptotes at points $X$ and $Y$. Prove that the segments $PX$ and $QY$ are equal.

The proof given requires an affine transformation such that the line becomes parallel to one of the axes of the hyperbola. However, as noted in the question, if the line is not parallel to one of the axes of symmetry of the hyperbola and intersects them both, then no affine transformation can make it parallel to one of the axes of the transformed hyperbola because the intersection points are also transformed.

Despite this, the result is true because the given proof is valid when it is interpreted properly. If the given line does intersect at least one of the axes of the hyperbola, then apply an affine transformation that fixes the asymptotes and makes the distance of $X$ and $Y$ from the other axis equal. In this case, the given line now becomes parallel to the other axis of the transformed hyperbola. Note the transformed axes of the original hyperbola need not be the same as the new axes of the transformed hyperbola, so there is no problem about parallel lines.

Compare this to the case of the center of a circle under inversion in a given circle where the inverted circle center is not the inverse of the center of the original circle unless this circle maps to itself.

The proof states

Then the desired equality would follow from symmetry.

which is easy to show and the proof is complete.

Perhaps an example will makes this clearer. Draw the hyperbola graph of $y=1/x$. Draw the line $y=x/2+1$ through the points $Y=(-2,0), X=(0,1), (2,2), (-4,1)$. Draw the symmetry axes $y=x$ and $y=-x$. Use the affine transformation $(x,y)\mapsto(x,2y)$. The new graph is the hyperbola $y=2/x$. The line is now $y=x+2$ which is parallel to the symmetry axis $y=x$ which is the same as the old one.

A proof using analytic geometry could be given. Let $a,b$ be nonzero numbers. Without loss of generality let the hyperbola be $y=1/x.$ Define

$$ d=\sqrt{c^2-4ab},\quad e=(c-d)/2,\quad f=(c+d)/2 $$

and points

$$ X=(0,c/b),\; Y=(c/a, 0),\; P = (e/a, f/b),\; Q = (f/a, e/b).$$

Verify that

$$ ef = ab,\; cP = fX+eY,\; cQ = eX+fY,\; P\!-\!X = Y\!-\!Q $$

to complete the proof.

Answer 2

It can make intersecting lines perpendicular. For take $M$ the middle of segment $XY$, $O$ the intersection of the asymptotes. In affine geometry we do not have angles, so we cannot say $OM \perp XY$. However, (assume the base field $\mathbb{R}$) we can choose a metric such that $OM \perp XY$. Then $OM$ is also an axis of symmetry of the pair of asymptotes. Therefore it is an axis of symmetry of the hyperbola, etc.

Notes:

1. In fact, we can make both $XY\perp OM$ and also the asymptotes perpendicular.

2. Perhaps one should prove that $\frac{M+N}{2}=\frac{X+Y}{2}$. This does not require the notion of length, and the field can be arbitrary, of characteristic not $2$. If that is to be allowed, then $Y-M = N-X$ would do.