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題1.

S

=1/2absinC

=1/2absinC(a2+b2-c2)/(2ab)/cosC

=tanC(a2+b2-c2)/4

BV1R4411T7hU

題2.

設(shè)漸近線與x軸夾角為θ

有tanθ=b/a

∠AOB=2θ

若A為直角頂點

其橫坐標(biāo)之比

即OA/OB

若b/a＜1

即b＜a

OA/OB

=cos(2θ)

=(1-tan2θ)/(1+tan2θ)

=(1-b2/a2)/(1+b2/a2)

=(a2-b2)/(a2+b2)

若b/a＞1

即b＞a

OA/OB

=cos(2θ)

=(1-tan2θ)/(1+tan2θ)

=(1-a2/b2)/(1+a2/b2)

=(b2-a2)/(a2+b2)

綜述

橫坐標(biāo)之比為

|a2-b2|/(a2+b2)

若B為直角頂點

則為其倒數(shù)

即(a2+b2)/|a2-b2|

故判定選項D