Given points P, Q on an ellipse a2x2+b2y2=1 (a>b>0), satisfying OP⊥OQ, the minimum of ∣OP∣×∣OQ∣ is ____.
Solution
Define P(∣OP∣cosθ,∣OP∣sinθ),Q(∣OQ∣cos(θ±2π),∣OQ∣sin(θ±2π)). We have ∣OP∣21=a2cos2θ+b2sin2θ,1◯ ∣OQ∣21=a2sin2θ+b2cos2θ.2◯ Then ∣OP∣21+∣OQ∣21=a21+b21. Therefore, ∣OP∣×∣OQ∣ reaches the minimum a2+b22a2b2 when ∣OP∣=∣OQ∣=a2+b22a2b2.
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Source: MathNet,
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