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latus rectum of the ellipse, which passes through $E$, and is always equal to the radius.

In order to account for this geometrically, we must observe, that rays proceeding from a point at an infinite distance from a mirror, are reflected to the principal focus, which is at the middle of the radius containing that point in its prolongation. Now $PQ$ being perpendicular to $AP$, a line drawn from $E$ to a point in $PQ$ infinitely distant from $P$ or $E$ must be perpendicular to $AE$, and the focus for rays proceeding from such a point will necessarily be $f',$ the middle point of the radius $EN'$ which is perpendicular to $AE$

It appears, that supposing the line $PQ$ to be infinitely extended both ways from $P$, and to be placed in $AE$ produced, at a distance from $E$ greater than half the radius, the image is a portion of an ellipse,^[1] extending from the extremity of the axis major to those of the latus rectum; we shall see hereafter how the ellipse may be supposed to be completed.

It is, however, necessary that we examine what change takes place in the image when $P$ is brought within the limit assigned above, namely, when $EP$ is not less than half the radius, or further when $P$ is placed on the other side of $E$.

In the first place, when $EP$ is half the radius, we have

$c={\frac {r}{2}};\quad \therefore e={\frac {r}{2c}}=1;\quad a={\frac {r}{2(1-1)}}=\infty .$

In this case then the ellipse changes to a parabola, (Fig. 40.)

Suppose now $EP$ be less than half the radius,

$c<{\frac {r}{2}};\quad \therefore e={\frac {r}{2c}}>1;\quad a=-{\frac {r}{2(e^{2}-1)}}=-{\frac {2c^{2}r}{r^{2}-4c^{2})}}.$

Here we have then a portion of an hyperbola.⁠(Fig. 41.)

When $P$ is at the centre, $c=0,\ e=\infty .$ The hyperbola becomes a straight line coincident with $PQ.$

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1. When $P$ is infinitely distant, the image is a semi-circle for $e{=}0.$