Find the eccentricity! -Part 1

Geometry Level pending

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

Point $O$ is the origin of the coordinate, $M$ is an arbitrary point on $C$ and above the $x$ axis, $H$ is a point on $MF_1$ .

Given that $MF_2 \perp F_1F_2$ , $MF_1 \perp OH$ , $|OH|=\lambda |OF_2|$ , where $\lambda \in (\dfrac{1}{3},\dfrac{1}{2})$ .

Find the range of the eccentricity of the hyperbola $C$ .

(\sqrt{2},2)

(\sqrt{2},\sqrt{3})

(1,\sqrt{3})

(1,\sqrt{2})

1 solution

David Vreken
Jan 2, 2020

Let $M$ have coordinates $(c, d)$ , so that $OF_2 = c$ and $MF_2 = d$ . Then $OF_1 = c$ and $OH = \lambda c$ , and by Pythagorean's Theorem, $HF_1 = c\sqrt{1-\lambda ^2}$ .

Since $\triangle OHF_1 \sim \triangle MF_2F_1$ by AA similarity, $\frac{\sqrt{1 - \lambda^2}}{\lambda} = \frac{2c}{d}$ .

Since $(c, d)$ is on the hyperbola, $\frac{c^2}{a^2} - \frac{d^2}{b^2} = 1$ .

By properties of a hyperbola, $a^2 + b^2 = c^2$ and $e = \frac{c}{a}$ .

These equations combine and rearrange to $e = \sqrt{\frac{1 + \lambda}{1 - \lambda}}$ , so that when $\lambda = \frac{1}{3}$ , $e = \sqrt{2}$ , and when $\lambda = \frac{1}{2}$ , $e = \sqrt{3}$ .

Since it is also an increasing function, the range of the eccentricity is $\boxed{(\sqrt{2}, \sqrt{3})}$ .

Find the eccentricity! -Part 1

1 solution

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