Page:Optics.djvu/60

 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, 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,

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

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 $$PQ.$$