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Geometry Difficulty 4.8 AIME Prove it China

Given points PP, QQ on an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a>b>0a > b > 0), satisfying OPOQOP \perp OQ, the minimum of OP×OQ|OP| \times |OQ| is ____.

Solution

Define
P(OPcosθ,OPsinθ),Q(OQcos(θ±π2),OQsin(θ±π2)). P(|OP| \cos \theta, |OP| \sin \theta), \\ Q(|OQ| \cos(\theta \pm \frac{\pi}{2}), |OQ| \sin(\theta \pm \frac{\pi}{2})).
We have
1OP2=cos2θa2+sin2θb2,1 \frac{1}{|OP|^2} = \frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}, \qquad \textcircled{1}
1OQ2=sin2θa2+cos2θb2.2 \frac{1}{|OQ|^2} = \frac{\sin^2\theta}{a^2} + \frac{\cos^2\theta}{b^2}. \qquad \textcircled{2}
Then
1OP2+1OQ2=1a2+1b2. \frac{1}{|OP|^2} + \frac{1}{|OQ|^2} = \frac{1}{a^2} + \frac{1}{b^2}.
Therefore, OP×OQ|OP| \times |OQ| reaches the minimum 2a2b2a2+b2\frac{2a^2b^2}{a^2+b^2} when OP=OQ=2a2b2a2+b2|OP| = |OQ| = \sqrt{\frac{2a^2b^2}{a^2+b^2}}.

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