I am a 12th student. I found this property in a reference book, without its proof.
So i tried to prove this myself but got stuck. Here's my attempt at the problem:
Basically the question is to prove $$\frac{1}{AC} + \frac{1}{AB} = \frac{2a}{b^2}$$
Where $\mathsf a$ and $\mathsf b$ are semi-major and semi-minor axes, $\mathsf CAB$ is the focal chord, $\mathsf A $ is the focus and $\mathsf AC$ and $\mathsf AB$ are its segments.
I took parametric form for $\mathsf B$ and $\mathsf C$ as:
$$\mathsf B (a\cos\alpha, b\sin\alpha ) \mathsf C (a\cos\beta, b\sin\beta )$$
Where $\alpha$ and $\beta$ are eccentric angles of $\mathsf B$ and $\mathsf C$.
I found the length of segments of focal chord as:
$$AB = a(1 -e\cos\alpha )$$ and $$AC = a(1 -e\cos\beta )$$ Where $A$ is $(ae, 0)$ and $e$ is the eccentricity.
After that i tried to apply harmonic mean but wasn't able to simplify the equation.
[Math] Prove that the semi Latus rectum of an ellipse is the harmonic mean of the segments of focal chord.
conic sectionscoordinate systems
Related Solutions
The polar form of the equation for an ellipse with "horizontal" semi-axis $a$ and "vertical" semi-axis $b$ is
$$r = \frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$$
Here, $\theta$ represents the angle measured from the horizontal axis ($30.5^\circ$ in your case), and $r$ is the distance from the center to the point in question (the radius you seek).
This exercise can be understood as an application of a general result about perimeter bisectors of triangles.
Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ meets $\overline{AB}$, then $$|\overline{CA}|+|\overline{AF^{\prime\prime}}| = |\overline{CB}|+|\overline{BF^{\prime\prime}}| \tag{$\star$}$$ so that $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$.
Proof of Proposition. Let the perpendicular to $\overline{FF^\prime}$ at $F^\prime$ meet the edges of the triangle at $A^\prime$ and $B^\prime$. By tangent properties of circles, we have $$\overline{CE}\cong\overline{CD} \qquad \overline{A^\prime E}\cong\overline{A^\prime F^\prime} \qquad \overline{B^\prime D}\cong\overline{B^\prime F^\prime}$$ Consequently, $|\overline{CA^\prime}| + |\overline{A^\prime F^\prime}| = |\overline{CB^\prime}| + |\overline{B^\prime F^\prime}|$, so that $\overline{CF}$ is a perimeter bisector of $\triangle A^\prime B^\prime C$. The Proposition holds by the similarity of $\triangle ABC$ and $\triangle A^\prime B^\prime C$. $\square$
The Proposition has a helpful corollary.
Corollary. Given $\triangle ABC$ with incenter $I$ and perimeter bisector $\overline{CF^{\prime\prime}}$, if $M$ is on $\overline{AB}$ such that $\overline{IM} \parallel \overline{CF^{\prime\prime}}$, then $M$ is the midpoint of $\overline{AB}$.
Proof of Corollary. The points of tangency of the triangle with its incircle separate the perimeter into three pairs of congruent segments, marked $a$, $b$, $c$. Thus, the semi-perimeter of $\triangle ABC$ is $a+b+c$, and since $|\overline{BC}| = b+c$, it follows that $|\overline{BF^{\prime\prime}}| = a = |\overline{AF}|$. Thus, $\overline{FF^{\prime\prime}}$ lies between congruent segments. In $\triangle FF^\prime F^{\prime\prime}$, segment $\overline{IM}$ passes through the midpoint of one side ($\overline{FF^\prime}$) and is parallel to another ($\overline{F^\prime F^{\prime\prime}}$); it necessarily meets the third side ($\overline{FF^{\prime\prime}}$) at its midpoint, which must also be the midpoint of $\overline{AB}$. $\square$
To solve the original problem, it basically suffices to embed the above triangle into an ellipse:
In the above, the ellipse's foci are $C$ and $F^{\prime\prime}$, and $\overline{AB}$ is a chord through the latter. The fundamental nature of ellipses implies that $(\star)$ holds; therefore, $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$. Moreover, the reflection property of ellipses implies that normals at $A$ and $B$ bisect angles $\angle CAF^{\prime\prime}$ and $\angle CBF^{\prime\prime}$; therefore, the intersection of these normals is the incenter of $\triangle ABC$. The result follows by the Corollary. $\square$
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Best Answer
Let $O$ be a focus of a non-circle conic, let $\overline{OR}$ be a semi-latus rectum, and let $\overline{PQ}$ be a focal chord not containing $R$. Let $O^\prime$, $P^\prime$, $Q^\prime$, $R^\prime$ be the projections of $O$, $P$, $Q$, $R$, respectively, onto the directrix associated with $O$. Note that, by the focus-directrix property of a conic with eccentricity $e\neq 0$, $$|OX|=e\,|XX^\prime| \quad\text{for}\quad X=P, Q, R \tag{1}$$
Defining $p:=|PP'|$, $q:=|QQ'|$, $r:=|RR'|$ (with $p\geq r \geq q$), we have, by obvious similar triangles in the figure,
proving the result. $\square$