椭圆X^2/a^2+y^2/b^2=1(a>b>0)上有两点A、B满足OA垂直于OB(O为坐标原点),求证:O到直线AB距离为定值第一问求出来1/OA^2+1/OB^2为定值,为(a^2+b^2)/(a^2b^2)
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![椭圆X^2/a^2+y^2/b^2=1(a>b>0)上有两点A、B满足OA垂直于OB(O为坐标原点),求证:O到直线AB距离为定值第一问求出来1/OA^2+1/OB^2为定值,为(a^2+b^2)/(a^2b^2)](/uploads/image/z/5285400-24-0.jpg?t=%E6%A4%AD%E5%9C%86X%5E2%2Fa%5E2%2By%5E2%2Fb%5E2%3D1%EF%BC%88a%3Eb%3E0%EF%BC%89%E4%B8%8A%E6%9C%89%E4%B8%A4%E7%82%B9A%E3%80%81B%E6%BB%A1%E8%B6%B3OA%E5%9E%82%E7%9B%B4%E4%BA%8EOB%EF%BC%88O%E4%B8%BA%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%EF%BC%89%2C%E6%B1%82%E8%AF%81%EF%BC%9AO%E5%88%B0%E7%9B%B4%E7%BA%BFAB%E8%B7%9D%E7%A6%BB%E4%B8%BA%E5%AE%9A%E5%80%BC%E7%AC%AC%E4%B8%80%E9%97%AE%E6%B1%82%E5%87%BA%E6%9D%A51%2FOA%5E2%2B1%2FOB%5E2%E4%B8%BA%E5%AE%9A%E5%80%BC%2C%E4%B8%BA%EF%BC%88a%5E2%2Bb%5E2%EF%BC%89%2F%28a%5E2b%5E2%29)
椭圆X^2/a^2+y^2/b^2=1(a>b>0)上有两点A、B满足OA垂直于OB(O为坐标原点),求证:O到直线AB距离为定值第一问求出来1/OA^2+1/OB^2为定值,为(a^2+b^2)/(a^2b^2)
椭圆X^2/a^2+y^2/b^2=1(a>b>0)上有两点A、B满足OA垂直于OB(O为坐标原点),求证:O到直线AB距离为定值
第一问求出来1/OA^2+1/OB^2为定值,为(a^2+b^2)/(a^2b^2)
椭圆X^2/a^2+y^2/b^2=1(a>b>0)上有两点A、B满足OA垂直于OB(O为坐标原点),求证:O到直线AB距离为定值第一问求出来1/OA^2+1/OB^2为定值,为(a^2+b^2)/(a^2b^2)
设A(x1,y1)B(x2,y2)
根据题意y1/x1*y2/x2=-1
即x1x2+y1y2=0
设MN方程:y=kx+m代入椭圆b²x²+a²y²=a²b²
整理:(a²k²+b²)x²+2kma²x+a²m²-a²b²=0
韦达定理:x1+x2=-2kma²/(a²k²+b²),x1*x2=(a²m²-a²b²)/(a²k²+b²)
y1y2=(kx1+m)(kx2+m)=k²x1x2+km(x1+x2)+m²
x1x2+k²x1x2+km(x1+x2)+m²=0
(a²m²-a²b²)/(a²k²+b²)+k²(a²m²-a²b²)/(a²k²+b²)-2k²m²a²/(a²k²+b²)+m²=0
化简:(a²+b²)m²=a²b²(1+k²)
m²/(1+k²)=a²b²/(a²+b²)
|m|/√(1+k²)=ab/√(a²+b²)
点O到直线MN的距离d=|m|/√(1+k²)=ab/√(a²+b²)为定值
1/OA²+1/OB²=(OA²+OB²)/(OA²*OB²)=AB²/(AB*d)²=1/d²=1/[a²b²/(a²+b²)]
=(a²+b²)/(a²b²)=1/a²+1/b²