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When $P$ is between $F$ and $A$, (Fig. 44.) $c>{\frac {r}{2}};\ e<1,$ and the image is part of an ellipse, namely, all but that part, which we found in the first case.

If we suppose $P$ to go outside of the circle beyond $A$, (Fig. 46.) we shall be led to the case of a convex mirror.

Our equation will still be

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

and the image will in all cases be a part of an ellipse, turning its convexity towards $P$.

When $P$ is on the circle at $A$, (Fig. 45.) the image extends from that point both ways to $f,\ f'$ the bisection of the radii $EN,\ EN',$ which are parallel to $PQ$.

When $P$ is at an infinite distance from $A$, the image is a semi-circle with centre $P$, and radius $EF$.

45.We may now show how the curve of the image, which we have in different cases found to be a part of a conic section, may be supposed to be completed.

Supposing in all cases the line to be infinite in extent each way.

In the first place, when $P$ is at an infinite distance, (Fig. 47.) the semi-circle $NAN'$ representing a concave mirror gives a semi-circular image $fFf';$ and the convex mirror represented by $NA'N'$ gives the image $fF'f'$, which completes the circle.

When $P$ is at a finite distance outside the circle, the concave and convex parts give together a complete ellipse $pfp'f'$, (Fig. 48.).

When $P$ is on the circle at $A'$, the ellipse is such as represented in Fig. 49, where $Ep$ is two-thirds of $EF.$

When $P$ is between $A'$ and $F'$, the ellipse cuts the circle, (Fig. 50.)

When $P$ is at $F'$ the middle point of $EA'$, (Fig. 51.) the two parts of the circle divided by the line, unite to produce a complete parabola.