过坐标原点O作两条互相垂直的射线,与椭圆C:3x2+4y2=12分别交与A,B两点,证明点O到直线A

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设过坐标原点O作的两条互相垂直的射线的方程分别为y=kx和y=(-1/k)x
并令它们与椭圆的交点分别为A(x1,y1)和B(x2,y2),则y1=kx1,y2=(-1/k)x2
将它们分别代入椭圆3x^2+4y^2=12中,得：(3+4k^2)*(x1)^2=12,(4+3k^2)*(x2)^2=12k^2
所以(x1)^2=12/(3+4k^2),(x2)^2=12k^2/(4+3k^2)
所以(|AB|)^2=(x2-x1)^2+(y2-y1)^2=(x2-x1)^2+{[(-1/k)x2]-kx1}^2
=(x2-x1)^2+[(1/k)x2+kx1]^2
=(x2)^2+(x1)^2-2x2x1+(1/k^2)*(x2)^2+(kx1)^2+2x2x1
=[(1+k^2)/k^2]*(x2)^2+(1+k^2)*(x1)^2
=[(1+k^2)/k^2]*[12k^2/(4+3k^2)]+(1+k^2)*[12/(3+4k^2)]
=[12(1+k^2)/(4+3k^2)]+[12(1+k^2)/(3+4k^2)]
=[12(1+k^2)]*(4+3k^2+3+4k^2)/[(4+3k^2)(3+4k^2)]
=[84(1+k^2)^2]/(12k^4+24k^2+12)
=[84(1+k^2)^2]/[12(1+k^2)^2]
=7而(|OA|)^2=(x1)^2+(y1)^2=(1+k^2)*(x1)^2,(|OB|)^2=(x2)^2+(y2)^2=[(1+k^2)/k^2]*(x2)^2
所以(|OA|*|OB|)^2=(1+k^2)*[12/(3+4k^2)]*[(1+k^2)/k^2]*[12k^2/(4+3k^2)]
=144(1+k^2)^2/[(3+4k^2)(4+3k^2)]
=144(1+k^2)^2/[12(1+k^2)^2]
=12点O到直线AB的距离为d,则S△OAB=1/2*|AB|*d=1/2*|OA|*|OB|
所以d^2=(|OA|*|OB|)^2/(|AB|)^2=12/7,所以d=√(12/7)=(2√21)/7,为定值
所以点O到直线AB的距离为定值(2√21)/7