视频 BV1ZE411q7iR 与 BV1R4411T7hU 解析

BV1ZE411q7iR

题1.

S

=1/2absinC

=1/2absinC(a²+b²-c²)/(2ab)/cosC

=tanC(a²+b²-c²)/4

BV1R4411T7hU

题2.

设渐近线与x轴夹角为θ

有tanθ=b/a

∠AOB=2θ

若A为直角顶点

其横坐标之比

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)

综述

横坐标之比为

|a²-b²|/(a²+b²)

若B为直角顶点

则为其倒数

即(a²+b²)/|a²-b²|

故判定选项D