Bisectors and conic section

Geometry Level pending

$O$ is the center of the coordinate, $F_1,F_2$ are left and right focus points of hyperbola: $\dfrac{x^2}{2020}-\dfrac{y^2}{2019}=1$ .

Point $P$ is an arbitrary point on the hyperbola and ray $l_1$ is the angle bisector of $\angle F_1PF_2$ .

Line $l_2$ passes through point $F_1$ and $l_1 \perp l_2$ . $l_1$ and $l_2$ meet at point $H$ .

Then $|OH|$ is a constant. Submit $|OH|^2$ .

The answer is 2020.

1 solution

David Vreken
Jan 4, 2020

Let $E$ be the intersection of $F_1H$ and $PF_2$ .

Then $\triangle PF_1H \cong \triangle PGH$ by the ASA congruency theorem, which means $F_1H \cong GH$ and $PF_1 \cong PG$ .

By the properties of a hyperbola, $PF_1 - PF_2 = 2a$ , and since $PG = PF_1$ , $PF_1 - PF_2 = PG - PF_2 = GF_2 = 2a$ .

Since $F_1H \cong GH$ , $H$ is a midpoint of $F_1G$ . Since $O$ is the midpoint of $F_1F_2$ , $OH$ is a midsegment of $\triangle GF_1F_2$ , so that $OH = \frac{1}{2}GF_2 = a$ , which means $|OH|^2 = a^2$ .

Therefore for the hyperbola $\frac{x^2}{2020} - \frac{y^2}{2019} = 1$ , $a^2 = |OH|^2 = \boxed{2020}$ .

Bisectors and conic section

The answer is 2020.

1 solution

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