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Geometry Difficulty 5.4 AIME, harder Find the answer

13. In the plane rectangular coordinate system xOyx O y, the hyperbola C:x2a2y2b2=1C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 has its right focus at FF. A line ll passing through point FF intersects the hyperbola CC at points AA and BB. If OFAB=FAFBO F \cdot A B=F A \cdot F B, find the eccentricity ee of the hyperbola CC.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Consider the line l:{x=c+tcosα,y=tsinα,l:\left\{\begin{array}{l}x=c+t \cos \alpha, \\ y=t \sin \alpha,\end{array}\right. and substitute it into the hyperbola equation b2x2a2y2=a2b2b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2} to get
(b2cos2αa2sin2α)t2+2b2ctcosα+b2c2a2b2=0. Then OFAB=FAFBc4b4c2cos2α4b4(b2cos2αa2sin2α)b2cos2αa2sin2α=b4b2cos2αa2sin2αc2c2cos2α(b2cos2αa2sin2α)=b22ac=c2a2e22e1=0e=2+1. \begin{array}{l} \left(b^{2} \cos ^{2} \alpha-a^{2} \sin ^{2} \alpha\right) t^{2}+2 b^{2} c t \cos \alpha+b^{2} c^{2}-a^{2} b^{2}=0 . \\ \text { Then } O F \cdot A B=F A \cdot F B \Rightarrow c \cdot \frac{\sqrt{4 b^{4} c^{2} \cos ^{2} \alpha-4 b^{4}\left(b^{2} \cos ^{2} \alpha-a^{2} \sin ^{2} \alpha\right)}}{\left|b^{2} \cos ^{2} \alpha-a^{2} \sin ^{2} \alpha\right|}=\frac{b^{4}}{\left|b^{2} \cos ^{2} \alpha-a^{2} \sin ^{2} \alpha\right|} \\ \Rightarrow c \cdot 2 \sqrt{c^{2} \cos ^{2} \alpha-\left(b^{2} \cos ^{2} \alpha-a^{2} \sin ^{2} \alpha\right)}=b^{2} \Rightarrow 2 a c=c^{2}-a^{2} \Rightarrow e^{2}-2 e-1=0 \Rightarrow e=\sqrt{2}+1 . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.