推薦：我們易求得漸進(jìn)線方程：y=(+-)3x/2.令漸近線y=3x/2傾斜角為@,則tan@=3/2,sin2@=2sin@cos@=(2sin@cos@)/[(sin@)^2+(cos@)^2]=2tan@/[1+(tan@)^2]=12/13.不防令A(yù)(2t1,3t1)B(2t2,-3t2),得lOAl=(13^0.5)t1,lOBl=(13^0.5)t2,于是S三角形AOB=(1/2)lOAllOBlsin2@=(1/2)*(13t1t2)*(12/13)=6t1t2=27/4,得t1t2=9/8.由向量AP=2PB,得P[(2t1+2*2t2)/(1+2),(3t1+2(-3t2))/(1+2)],即P((2t1+4t2)/3,t1-2t2),第一問題我們知b^2=9a^2/4,于是雙曲線方程可寫為x^2-(4/9)y^2=a^2,P點(diǎn)在雙曲線上則有：[(2t1+4t2)/3]^2-(4/9)(t1-2t2)^2=a^2,即(t1+2t2)^2-(t1-t2)^2=(9/4)a^2得到8t1t2=(9/4)a^2,又t1t2=9/8,所以a^2=4,那么雙曲線方程為x^2/4-y^2/9=1