Let's have the line $k$ which is gonna be the bisector of the $F_2PF_2'$ angle. We do have the ellipse with foci $F_1$ and $F_2$ and a point $P$ on the ellipse curve. It is pretty obvious that for ellipse in one point there can be only one tangent. To do that, we're gonna bisect the angle with a line and prove that this line is the tangent by showing that there are no other points that belong to the ellipse on that line.
We're gonna prove the focal property by proving that a tangent bisects the given angle. $\theta_1=\theta_3$ by vertical angles, so So $L$ is the tangent line to $E$ at $P$. So $QF_2 = QF_2'$.īy the definition of ellipse, since $P$ and $Q$ are on $E$, $F_1Q + QF_2 = F_1P + PF_2$. Since $Q$ is on $L$, $Q$ is equidistant from $F_2$ and $F_2'$. Then $L$ meets $E$ at another point $Q$ on $E$. Suppose $L$ is not tangent to $E$ at $P$. $L$ is thus the perpendicular bisector of $F_2F_2'$, the mirror line for $F_2$ and $F_2'$.Ĭlaim: $L$ is the tangent line to $E$ at $P$. To prove: $F_1P$ and $F_2P$ make equal angles with the tangent line to $E$ at $P$.Įxtend $F_1P$ past $P$ to $F_2'$ such that $F_2P=PF_2'$. Take ellipse $E$ with foci at $F_1$ and $F_2$. Hopefully I have been very clear in my construction and argument.Ī ray of light starting at one focus will bounce off the ellipse and go through the other focus. This expands upon and clarifies the correct answers above by Abel, Jantomedes, and Peter. Thus, $t$ halves the angle $F_2XG$ and the normal halves the angle $F_2XF_1$. Proof: all $Q\neq X$ and $Q\in t$ do not lie on the ellipse:
Lengthen $F_1X$ by $F_2X$, find $G$, the axis $t$ of $F_2G$ is tangent to the ellipse through $X$. This proof is from Omar Antolín Camarena : īelow is the geometrical construction from Planungen11.PDF This proves that cos(angles) are opposite, and relevant angles are equal. We receive, after not much calculation, with n as a vector in the direction of the tangent, (1) |P-F1|+|P-F2|=const, which is a definition of ellipse.Īssume P is parametrized as P(t), and differentiate by t. I present it in short in case the original source disappears. There is also an easy analytical proof of the reflection property. I add the relevant page below, in case the original source disappears, and author is probably Monika Schwarze. Here is a better presentation for it, on page 4, (in german), with a clear drawing, at. The answer of abel and Jantomedes above is the correct geometrical and easy construction, short of drawings.
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