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China Mathematical Competition

China geometry

Problem

Given points

P

,

Q

on an ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

(

a > b > 0

), satisfying

O P ⊥ O Q

, the minimum of

∣ O P ∣ \times ∣ O Q ∣

is ____.

Solution

Define

P (∣ O P ∣ cos θ, ∣ O P ∣ sin θ), Q (∣ O Q ∣ cos (θ \pm \frac{π}{2}), ∣ O Q ∣ sin (θ \pm \frac{π}{2})) .

We have

\frac{1}{∣ O P ∣ ^{2}} = \frac{cos ^{2} θ}{a ^{2}} + \frac{sin ^{2} θ}{b ^{2}}, \overset{◯}{1}

\frac{1}{∣ O Q ∣ ^{2}} = \frac{sin ^{2} θ}{a ^{2}} + \frac{cos ^{2} θ}{b ^{2}} . \overset{◯}{2}

Then

\frac{1}{∣ O P ∣ ^{2}} + \frac{1}{∣ O Q ∣ ^{2}} = \frac{1}{a ^{2}} + \frac{1}{b ^{2}} .

Therefore,

∣ O P ∣ \times ∣ O Q ∣

reaches the minimum

\frac{2 a ^{2} b ^{2}}{a ^{2} + b ^{2}}

when

∣ O P ∣ = ∣ O Q ∣ = \frac{2 a ^{2} b ^{2}}{a ^{2} + b ^{2}}

.

Final answer

2a^2b^2/(a^2+b^2)

Techniques

Cartesian coordinatesTrigonometryQM-AM-GM-HM / Power Mean