PA + PB = OM + ON in an hyperpbola

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PA + PB = OM + ON in an hyperpbola

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Source: Mathematical Reflections U237

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#1 h Jun 6, 2020, 10:24 AM

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Let $H$ be a hyperbola with foci $A$ and $B$ and center $O$ . Let $P$ be an arbitrary point on $H$ and let the tangent of $H$ through $P$ cut its asymptotes at $M$ and $N$ Prove that $PA + PB = OM + ON$

by Luis Gonzalez, Maracaibo, Venezuela

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#2 h Apr 6, 2022, 6:44 PM

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Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with point $P(\frac{a}{\cos \alpha},b\tan \alpha)$ .

Foci $A(c,0)$ anf $B(-c,0)$ with $c=\sqrt{a^{2}+b^{2}}$ .
The tangent line $b \cdot x-a\sin \alpha \cdot y=ab\cos \alpha$ cuts the asymptote $y=\frac{b}{a}x$ in the point $M(\frac{a\cos \alpha}{1-\sin \alpha},\frac{b\cos \alpha}{1-\sin \alpha})$
and the asymptote $y=-\frac{b}{a}x$ in the point $N(\frac{a\cos \alpha}{1+\sin \alpha},\frac{b\cos \alpha}{1+\sin \alpha})$ .

Then $OM=\frac{\sqrt{a^{2}+b^{2}}\cos \alpha}{1-\sin \alpha}=\frac{c\cos \alpha}{1-\sin \alpha}$ and $ON=\frac{c\cos \alpha}{1+\sin \alpha}$ .
$OM+ON=\frac{2c}{\cos \alpha}$ .
$PA=\frac{c}{\cos \alpha}-a$ and $PB=\frac{c}{\cos \alpha}+a$ , then $PA+PB=\frac{2c}{\cos \alpha}$ .

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#3 h Apr 7, 2022, 1:05 PM • 1 Y

Y by old_csk_mo

Nice problem.

Claim. $AMBN$ is cyclic.
Proof. Let $M',N'$ be reflections of $M,N$ over $O.$ We obtain $\angle MBN=$ $\pi -\angle M'BN'=$ $\pi -\angle MAN$ as desired.

Now let $K,C$ be reflections of $M,B$ wrt perpendicular bisectors of $AB,MN$ respectively. By properties of hyperbola $O,P$ bisect $AB,MN$ respectively, $PN,OB$ bisect angles $APB,MON,$ so $\overline{K-O-N},\overline{C-P-A}$ . But $A,K,M,B,C,N$ are concyclic and $|AK|=|BM|=|CN|,$ therefore $|PA|+|PB|=|AC|=|KN|=|OM|+|ON|.$

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