Page:Optics.djvu/59

 Then supposing the object $$PQ$$ to be beyond $$E$$ the centre of the mirror,

Let $$r$$ represent the radius $$EA,$$

$$\theta$$ angle $$PEQ,$$

$$c$$ line $$EP,$$

$$q \Bigl( = \frac{c}{\cos \theta} \Bigr)$$ $$EQ,$$

$$q'$$ $$.

Then, referring to page 8, we find $$\frac{1}{q'} = \frac{2}{r} +\frac{1}{q} =\frac{2}{r} +\frac{1}{c} \cos \theta.$$

Now the polar equation to an ellipse referred to the near focus is

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

which coincides with the above, provided $$a (1-e^2) = \frac{r}{2},$$ and

$$\frac{a(1-e^2)}{e} = c; \quad \mathrm{that~is,~} e = \frac{r}{2c},$$

and $$ a = \frac{r}{2 (1-e^2)} = \frac{r}{2 \Bigl(1 - \frac{r^2}{4 c^2}\Bigr)} = \frac{2 c^2 r}{4 c^2 - r^2}.$$

It is also necessary, that $$e$$ be less than unity, that is, $$c>\frac{r}{2}$$.

44.There are some things to be remarked in this elliptic image. We found the quantity $$a \ (1-e^2)$$ to be constantly equal to the half radius. Now $$a \ (1-e^2)$$ is half the latus rectum of the ellipse, which is also the radius of curvature at the vertex. It appears then, that although the place of the vertex $$p,$$ and the magnitude of the ellipse, depend on the place of the point $$P,$$ yet the curvature of the image at the vertex is invariable, as is the

Errata