Question

椭圆x²/4+y²/2=1上有两点A,B,O为坐标原点,若直线OA,OB的斜率乘积为-1／2,求证：

Answer 1

设OA斜率k1=m

k1k2=-1/2

OB斜率k2=(-1/2)/k1=-1/(2m)

OA所在直线方程y=k1x=mx,代入x²/4+y²/2=1：

x^2/4+(mx)^2/2=1

(2m^2+1)x^2=4

x^2=4/(2m^2+1)

y^2=m^2x^2=4m^2/(2m^2+1)

|OA|^2=4/(2m^2+1)+4m^2/(2m^2+1)=4(m^2+1)/(2m^2+1)

OB所在直线方程y=k2x=-1/(2m) x,代入x²/4+y²/2=1：

x^2/4+x^2/（8m^2)=1

(2m^2+1)x^2=8m^2

x^2=8m^2/(2m^2+1)

y^2=x^2/(4m^2)=2/(2m^2+1)

|OB|^2=8m^2/(2m^2+1)+2/(2m^2+1)=2(4m^2+1)/(2m^2+1)

|OA|^2+|OB|^2=4(m^2+1)/(2m^2+1) + 2(4m^2+1)/(2m^2+1)

=2(2m^2+2+4m^2+1)/(2m^2+1)

=6(2m^2+1)/(2m^2+1)

=6