Find the eccentricity! - Part 3

Geometry Level pending

On the coordinate plane, point A A is on the ellipse: x 2 a 2 + y 2 b 2 = 1 ( a > b > 0 ) \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (a>b>0) . Point F F is the right focus point of the ellipse.

Point A , B A,B are symmetry about the origin point O O , and A F ⊥ B F AF \perp BF .

e e is the eccentricity of the ellipse. If ∠ A B F \angle ABF ranges from [ π 6 , π 4 ] [\dfrac{\pi}{6},\dfrac{\pi}{4}] , then find the range of e e .

[ 2 2 , 1 ) [\dfrac{\sqrt{2}}{2},1) [ 2 2 , 3 − 1 ] [\dfrac{\sqrt{2}}{2},\sqrt{3}-1] [ 3 3 , 6 3 ] [\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{6}}{3}] [ 2 2 , 3 2 ] [\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{3}}{2}]

Alice Smith by Alice Smith , 20, China

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1 solution

Let the coordinates of A A be ( h , k ) (h, k) . Then those of B B are ( − h , − k ) (-h, -k) . Eccentricity of the ellipse is e = 1 − ( b a ) 2 e=\sqrt {1-(\dfrac{b}{a})^2} . From these and the condition of perpendicularity of A F ‾ \overline {AF} and B F ‾ \overline {BF} we get h = a e 2 e 2 − 1 , k = a ( 1 − e 2 ) e h=\dfrac{a}{e}\sqrt {2e^2-1}, k=\dfrac{a(1-e^2)}{e} . In the lower limit of the given range, thus, e = 2 3 + 1 = 3 − 1 e=\dfrac{2}{\sqrt 3+1}=\sqrt 3-1 . In the upper limit, we get e = 1 2 = 2 2 e=\dfrac{1}{\sqrt 2}=\dfrac{\sqrt 2}{2} . Hence e e has the range [ 2 2 , 3 − 1 ] [\dfrac{\sqrt 2}{2}, \sqrt 3-1]

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