Page:Optics.djvu/103

 which agrees with that above, if

If $y^{2}$ be less than unity, the integrated equation becomes

which agrees with the equation to the ellipse,

The general equation integrated gives

The curve to which this belongs is a certain oval, which Descartes has described.

105. Newton's second proposition is:

To find the form of a convex lens, that shall refract light accurately from one point to another.

He supposes the first surface given, and determines the second thus, (Fig. 102.)

Let $=(e^{2}−1) {(x−ae)^{2}−a^{2}}|undefined$ be the focus of incident, $=(e^{2}−1) {x^{2}−2aex+(e^{2}−1)a^{2}}|undefined$ of refracted rays; $m=e$ the course of a ray; $mn⁄m^{2}−1=$, $−ae$, circular arcs with centers $∴ a=−n⁄m^{2}−1$, $n^{2}⁄m^{2}−1=$; $(e^{2}−1)a^{2}$, $a^{2}=n^{2}⁄(m^{2}−1)^{2}$, orthogonal trajectories to $m$.

Let $y^{2}=n^{2}+2mnx−(1−m^{2})x^{2}$, $y^{2}$, $=(1−e^{2}) {a^{2}−(x−ae)^{2}}|undefined$, be produced so that $y^{2}$, $=(1−e^{2}) {(1−e^{2})a^{2}+2aex−x^{2}}|undefined$; $u+mv=n$. Join $√x^{2}+y^{2}+m√(c−x)^{2}+y^{2}=n$, and let a circle with centre $A$, and radius $B$ cut it in $ADFB$. Draw $CP$ parallel to $ER$. In the first place $A$; $B$. For let $CQ$, $ES$