Find the eccentricity! - Part 4

Geometry Level pending

On a coordinate plane, ellipse $C_1:\dfrac{x^2}{a^2_1}+\dfrac{y^2}{b^2_1}=1 (a_1>b_1>0)$ and hyperbola $C_2:\dfrac{x^2}{a^2_2}-\dfrac{y^2}{b^2_2}=1 (a_2,b_2>0)$ has the same focus point $F_1,F_2$ .

Point $P$ is the intersection point of $C_1$ and $C_2$ in the first quadrant, and $|F_1F_2|=2|PF_2|$ .

$e_1$ is the eccentricity of $C_1$ and $e_2$ is the eccentricity of $C_2$ . Find the range of $e_2-e_1$ .

(\dfrac{1}{3},\infty)

(\dfrac{1}{2},\infty)

[\dfrac{1}{3},\infty)

[\dfrac{1}{2},\infty)

1 solution

David Vreken
Jan 3, 2020

By the properties of an ellipse, $|PF_1| + |PF_2| = 2a_1$ , and by the properties of a hyperbola, $|PF_1| - |PF_2| = 2a_2$ . These equations combine to $|PF_1| = a_1 + a_2$ and $|PF_2| = a_1 - a_2$ . Also by the properties of an ellipse and hyperbola, $|F_1F_2| = 2a_1e_1 = 2a_2e_2$ .

Since $|F_1F_2| = 2|PF_2|$ , $2a_1e_1 = 2(a_1 - a_2)$ . Substituting $a_2 = \frac{a_1e_1}{e_2}$ and rearranging eliminates $a_1$ and gives $e_2 = \frac{e_1}{1 - e_1}$ .

For a hyperbola, $e_2 > 1$ , which means $\frac{e_1}{1 - e_1} > 1$ , which solves to $e_1 > \frac{1}{2}$ . For an ellipse, $e_1 < 1$ . Therefore, $\frac{1}{2} < e_1 < 1$ .

Therefore, $D = e_2 - e_1 = \frac{e_1}{1 - e_1} - e_1 = \frac{e_1^2}{1 - e_1}$ . For $\frac{1}{2} < e_1 < 1$ , the range of $D$ is $\boxed{(\frac{1}{2}, \infty)}$ .

Find the eccentricity! - Part 4

1 solution

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