Find the asymptote! - Part 1

Geometry Level pending

On the coordinate, point $F_1,F_2$ are left and right focus points of the hyperbola $C:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 (a>0,b>0)$ respectively.

Line $l$ passes through $F_1$ and is the tangent to circle $x^2+y^2=a^2$ . $l$ intersects with the right half of the hyperbola at point $M$ .

If $\angle F_1MF_2=\dfrac{\pi}{4}$ , find the equation of the asymptotes of the hyperbola $C$ .

y=\pm x

y=\pm \sqrt{2} x

y=\pm 2x

y=\pm \sqrt{3} x

1 solution

David Vreken
Jan 3, 2020

Let $O$ be the origin, $T$ be the tangent point, and $\theta = \angle TF_1O$ .

Then by the properties of a circle, $OT = a$ , and by the properties of a hyperbola, $OF_1 = OF_2 = ae$ and $MF_1 - MF_2 = 2a$ . Let $MF_2 = p$ . Then $MF_1 = 2a + p$ .

By the law of sines on $\triangle MF_1F_2$ , $\frac{\sin \theta}{p} = \frac{\sin \frac{\pi}{4}}{2ae}$ . Substituting $\sin \theta = \frac{a}{ae}$ from $\triangle TOF_1$ and simplifying gives $p = 2\sqrt{2}a$ .

Now by the law of cosines on $\triangle MF_1F_2$ , $(2ae)^2 = (2a + 2\sqrt{2}a)^2 + (2\sqrt{2}a)^2 - 2(2a + 2\sqrt{2}a)(2\sqrt{2}a)\cos \frac{\pi}{4}$ , which simplifies to $e = \sqrt{3}$ .

The asymptotes of a hyperbola are $y = \pm \frac{b}{a}x$ . In this case, $\frac{b}{a} = \sqrt{e^2 - 1} = \sqrt{(\sqrt{3})^2 - 1} = \sqrt{2}$ . Therefore, the asymptotes of this hyperbola are $\boxed{y = \pm \sqrt{2}x}$ .

Find the asymptote! - Part 1

1 solution

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