Hyperbola and the angle bisector

Eccentricity of the hyperbola is $\sqrt {1+3}=2$ . So the coordinates of $F_1$ and $F_2$ are $(-2,0)$ and $(2,0)$ respectively. Let the coordinates of $P$ and $A$ be $(h, k)$ and $(a, 0)$ respectively. Then slopes of the line segments $\overline {PF_1}, \overline {PF_2}$ and $\overline {PA}$ are $\dfrac{k}{h+2}, \dfrac{k}{h-2}$ and $\dfrac{k}{h-a}$ respectively. So we have

$h^2-\dfrac {k^2}{3}=1$

$\tan \dfrac{2π}{3}=-\sqrt 3=\dfrac{\dfrac{k}{h-2}-\dfrac{k}{h+2}}{1+\dfrac{k^2}{h^2-4}}$

$\tan \dfrac{π}{3}=\sqrt 3=\dfrac{\dfrac{k}{h-a}-\dfrac{k}{h+2}}{1+\dfrac{k^2}{(h-a) (h+2)}}$ .

From the first two equations we get $h=\dfrac{\sqrt 5}{2}, k=\dfrac {\sqrt 3}{2}$ .

Substituting the values of $h$ and $k$ in the third equation we get $a=\dfrac{2}{\sqrt 5}$ and $|\overline {PA}|=\sqrt {(h-a) ^2+k^2}=\dfrac {2}{\sqrt 5}=\boxed {\dfrac {2\sqrt 5}{5}}$ .

Let the $x$ -coordinate of $P$ be $p$ . Since $P$ is on the hyperbola $x^2 - \frac{y^2}{3} = 1$ , its $y$ -coordinate is $\sqrt{3p^2 - 3}$ .

Let $B$ be the intersection of the line through $P$ perpendicular to $F_1F_2$ . Then $BF_1 = p + 2$ , $BF_2 = 2 - p$ , and $BP = \sqrt{3p^2 - 3}$ , and by the Pythagorean theorem on $\triangle PBF_1$ and $\triangle PBF_2$ , $PF_1 = \sqrt{(p + 2)^2 + 3p^2 - 3} = 2p + 1$ and $PF_2 = \sqrt{(2 - p)^2 + 3p^2 - 3} = 2p - 1$ .

By the law of cosines on $\triangle PF_1F_2$ , $4^2 = (2p + 1)^2 + (2p - 1)^2 - 2(2p + 1)(2p + 1) \cos \frac{2\pi}{3}$ , which solves to $p = \frac{\sqrt{5}}{2}$ for $p > 0$ . Therefore, $PF_1 = 2(\frac{\sqrt{5}}{2}) + 1 = \sqrt{5} + 1$ and $PF_2 = 2(\frac{\sqrt{5}}{2}) - 1 = \sqrt{5} - 1$ .

The length of an angle bisector $d$ of a triangle is $d^2 = bc - \frac{a^2bc}{(b + c)^2}$ , so $|PA| = \sqrt{(\sqrt{5} + 1)(\sqrt{5} - 1) - \frac{4^2(\sqrt{5} + 1)(\sqrt{5} - 1)}{(\sqrt{5} + 1 + \sqrt{5} - 1)^2}} = \boxed{\frac{2\sqrt{5}}{5}}$ .

It should be noted that the angle bisector $PA$ of $\angle F_1PF_2$ is actually tangential to the hyperbola at $P$ (see note). For hyperbola $\dfrac {x^2}{a^2} - \dfrac {y^2}{b^2} = 1$ , the tangent at $P(x_0, y_0)$ is given by $\dfrac {x_0}{a^2}x - \dfrac {y_0}{b^2}y = 1$ . For $A(x_A, 0)$ , $\implies x_A =\dfrac {a^2}{x_0}$ . Then $|PA| = \sqrt{\left(x_0-\dfrac {a^2}{x_0}\right)^2+y_0^2}$ .

To find $x_0$ and $y_0$ , let $F_1(-c,0)$ and $F_2(c,0)$ . Then

$\begin{aligned} \frac {\frac {c+x_0}y_0 + \frac {c-x_0}y_0}{1-\frac {c^2-x_0^2}{y_0^2}} & = \tan \frac {2\pi}3 \\ \frac {2cy_0}{y_0^2-c^2+x_0^2} & = - \sqrt 3 \\ \frac {2cy_0}{y_0^2 - c^2 + \frac {a^2}{b^2}y_0^2 + a^2} & = - \sqrt 3 & \small \blue{\text{Putting }a=1, b = \sqrt 3, c=2} \\ 4\sqrt 3 y_0^2 + 12y_0 - 9\sqrt 3 & = 0 \\ \implies y_0 & = \frac {\sqrt 3}2 \\ \implies x_0 & = \sqrt{1+\frac {y_0^2}3} = \frac {\sqrt 5}2 \\ \implies |PA| & = \sqrt{\left(x_0-\frac 1{x_0}\right)^2+y_0^2} = \boxed{\frac {2\sqrt 5}5} \end{aligned}$

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Note: I am not able to come up with a general proof but we can see that when $P$ is at the vertex $(a,0)$ , $\angle F_2PF_2 = \pi$ and the tangent is vertical. And when $P(\infty, \infty)$ , the tangent is parallel to the asymptote.

Let $P(x,y)$ be its coordinate, then we have $x^2-\frac{y^2}3=1\implies y^2=3x^2-3.$

Since $\angle F_1PF_2=2\pi/3$ , using Cosine Law we have $16=F_1F_2^2=PF_1^2+PF_2^2-2PF_1\cdot PF_2\cos\bigg(\dfrac{2\pi}3\bigg)=PF_1^2+PF_2^2+PF_1\cdot PF_2.$

By $PF_1^2=(x+2)^2+y^2, \quad PF_2^2=(x-2)^2+y^2,$

we have $\begin{aligned}16&=(x+2)^2+y^2+(x-2)^2+y^2+\sqrt{(x+2)^2+y^2}\sqrt{(x-2)^2+y^2}\\ &=2x^2+8+2y^2+\sqrt{(x+2)^2+3x^2-3}\sqrt{(x-2)^2+3x^2-3}\\ &=2x^2+8+2(3x^2-3)+\sqrt{4x^2+4x+1}\sqrt{4x^2-4x+1}\\ &=8x^2+2+(2x+1)(2x-1)\quad\text{ because }x>1\\ &=12x^2+1\\ \therefore x&=\dfrac{\sqrt5}2.\end{aligned}$

It is pretty lucky that the terms in square root is a perfect square! By apply $y^2=3x^2-3$ we find $y=\sqrt3/2$ .

From this we find $PF_1=\sqrt 5+1, PF_2=\sqrt5-1$ . Since $PA$ is the angle bisector of $\angle F_1PF_2$ , using area argument we find $\dfrac{PF_1}{PF_2}=\dfrac{AF_1}{AF_2},$

Using $AF_2$ as the sole variable here, we find $AF_2=\dfrac{10-2\sqrt 5}{5}$ . Therefore the $x$ -coordinate of $A$ is $2-\dfrac{10-2\sqrt 5}{5}=\dfrac{2\sqrt5}{5},$

Using $P(\sqrt 5/2, \sqrt3/2), A(2\sqrt5/5, 0)$ , we get $PA=\boxed{\dfrac{2\sqrt5}{5}}$ .

Note: This also shows that $\Delta APF_2$ is equilateral.