Question

过椭圆c:(x²/4)+y²=1的右焦点的直线l交椭圆于A,B两点，且|AB|=3/2。

过椭圆c:(x²/4)+y²=1的右焦点的直线l交椭圆于A,B两点，且|AB|=3/2。求直线l的斜率

Answer 1

椭圆C:x²/a²+y²/b²=1（a>b>0)过点P(2,1),∴4/a^2+1/b^2=1,①离心率e=c/a=√3/2，∴a^2=4c^2/3,b^2=c^2/3,代入①，6/c^2=1,c^2=6,∴a^2=8,b^2=2,椭圆C的方程是x^2/8+y^2/2=1.②设L:y=kx+m,③代入②*8，得x^2+4(k^2x^2+2kmx+m^2)=8,整理得(1+4k^2)x^2+8kmx+4m^2-8=0,设A(x1,y1),B(x2,y2),则x1+x2=-8km/(1+4k^2),x1x2=(4m^2-8)/(1+4k^2),向量PA*PB=(x1-2,y1-1)*(x2-2,y2-1)=(x1-2)(x2-2)+(kx1+m-1)(kx2+m-1)(由③)=(1+k^2)x1x2+[k(m-1)-2](x1+x2)+4+(m-1)^2=[(1+k^2)(4m^2-8)-8km(km-k-2)]/(1+4k^2)+4+(m-1)^2=0,∴(1+k^2)(4m^2-8)-8km(km-k-2)+(1+4k^2)[4+(m-1)^2]=0,整理得4k^2*[m^2-2-2m^2+2m+4+(m-1)^2]+16km+4m^2-8+4+(m-1)^2=0,12k^2+16km+5m^2-2m-3=0,解得k=(1-m)/2,或k=-(5m+3)/6,∴L:y=(1-m)x/2+m,（A,B均异于P,舍)或y=-(5m+3)x/6+m,过定点(6/5,-3/5).证完.