Page:Optics.djvu/61

37

When $P$ is between $A$ and $E$, $EP$ or $c$ is negative; our formula then becomes

${\frac {1}{q'}}={\frac {2}{r}}-{\frac {1}{c}}\cos \theta ,$

which answers to

${\frac {1}{\rho }}={\frac {1}{a(1-e^{2})}}-{\frac {e}{a(1-e^{2})}}\cos \theta ,$

or ${\frac {1}{\rho }}=-{\frac {1}{a(e^{2}-1)}}+{\frac {e}{a(e^{2}-1)}}\cos \theta ,$

the former of which is the equation to an ellipse, when the angle $\theta$ is measured from the farther vertex, and is consequently the supplement of that used in the former cases; the latter is the equation to a pair of hyperbolas.

This latter case we will now examine, as it comes first in order.

Let then $P,$ (Fig. 42.) be between $E$ and $F,$ (the principal focus). In the first place, we know that if $k$ be that point in $PQ$ for which $Ek=EF,$ the image of $K$ must be infinitely distant, so that the line $Ek$ must be parallel to one of the asymptotes.

Again, if $g$ be the point where $PQ$ cuts the circle, its image coincides with it. Every point between $k$ and $g$ has its image without the circle, the distances of these images diminishing gradually from infinity to the radius.

It appears then that the image in this case consists of an hyperbola, and its conjugate wanting the part between the vertex and the extremities of the latus rectum.

When $P$ coincides with $F,$ (Fig. 43.) the image is a parabola, wanting the part about the vertex extending to the extremities of the latus rectum.

The equation in this case takes the form

${\frac {1}{q'}}={\frac {2}{r}}-{\frac {2}{r}}\cos \theta ={\frac {2}{r}}(1-\cos \theta ),$

or $q'={\frac {{\frac {1}{2}}r}{1-\cos \theta }},$

which we know to be that of the parabola, in which the angle $\theta$ is measured from the axis, not beginning at the vertex.