COURSE OF PROJECTIVE GEOMETRY

§ 4: answers

*O12* The line at infinity in α corresponds under the rotation to the line at infinity in β (indeed, a bundle of parallel lines is mapped onto a bundle of parallel lines).

The line at infinity in β is projected onto *u* (a bundle of parallel lines in β is mapped onto a bundle through a point of *u*).

Under the central projection, the line at infinity of α corresponds to *s*. So the original is *s"*, that is the line in α with which *s* coincides after rotating β onto
α (we have to agree upon a direction of rotating).

*O13* Choose a point *P"* with image *P'* (unequal to *P"*) and a point *Q"*, not on *P"**P'*, with image *Q'* (unequal to *Q"*). Let
*V* be the point of intersection *P"**P'*.*Q"**Q'*.

Now let *X"* be a point, not *V*, not on *t* or *P"**P'* or *Q"**Q'*.

Since a line and its image intersect on *t*, the triangles *P"**Q"**X"* and *P'**Q'**X'* are line perspective from *t*. So they are correspondingly
point perspective. So *X"**X'* goes through *V*.

Next, continue with *X"* on *P"**P'*, etcetera.

*O14*

*Q'* lies on *V**Q"*, but also on *P'**S* where *S* = *t*.*P"**Q"*.

*O15*

The construction is the same as in *O14*.

Start with a line *l*, unequal to *t* or *P'**P"*, and the point at infinity *Q'* of that line *l*.

Let *T* be the intersection point *t*.*P'**Q'*, so that *Q"* is lying on *P"**T*.

Then we find *Q"* as the intersection point of *P"**T* and *V**Q'*.

*O16* Choose *X"* on *c"*.*X'* lies on *V**X"*, but also on *P'**Y'*, where *Y"* = *P"**X"*.*s"*.

Since *Y"* lies on *s"*, *Y'* is the point at infinity of *V**Y"*.

We have three cases: *s"* and *c"* have two, one or zero intersection points.
Then *c'* becomes, respectively, a hyperbola, parabola or ellipse. We see this happen if we construct many pairs (*X"*,*X '* ), with *X"* on *c"*.