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:|OA|²+
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解决时间 2021-11-12 16:30
- 提问者网友:记得曾经
- 2021-11-12 07:36
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:|OA|²+
最佳答案
- 五星知识达人网友:你可爱的野爹
- 2021-11-12 09:07
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
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
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