椭圆X^2/25+Y^2/9=1上不同三点A(x1,y1),B(4,9/5)C(x2,y2)与焦点F(4,0)的距离成等差数列(1) 求证X1+x2=8(2) 若线段AC的垂直平分线与x周的焦点的交点为T,求直线BT的斜率k
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![椭圆X^2/25+Y^2/9=1上不同三点A(x1,y1),B(4,9/5)C(x2,y2)与焦点F(4,0)的距离成等差数列(1) 求证X1+x2=8(2) 若线段AC的垂直平分线与x周的焦点的交点为T,求直线BT的斜率k](/uploads/image/z/3951127-55-7.jpg?t=%E6%A4%AD%E5%9C%86X%5E2%2F25%2BY%5E2%2F9%3D1%E4%B8%8A%E4%B8%8D%E5%90%8C%E4%B8%89%E7%82%B9A%28x1%2Cy1%29%2CB%284%2C9%2F5%29C%28x2%2Cy2%29%E4%B8%8E%E7%84%A6%E7%82%B9F%284%2C0%29%E7%9A%84%E8%B7%9D%E7%A6%BB%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%281%29+%E6%B1%82%E8%AF%81X1%2Bx2%3D8%282%29+%E8%8B%A5%E7%BA%BF%E6%AE%B5AC%E7%9A%84%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BF%E4%B8%8Ex%E5%91%A8%E7%9A%84%E7%84%A6%E7%82%B9%E7%9A%84%E4%BA%A4%E7%82%B9%E4%B8%BAT%2C%E6%B1%82%E7%9B%B4%E7%BA%BFBT%E7%9A%84%E6%96%9C%E7%8E%87k)
椭圆X^2/25+Y^2/9=1上不同三点A(x1,y1),B(4,9/5)C(x2,y2)与焦点F(4,0)的距离成等差数列(1) 求证X1+x2=8(2) 若线段AC的垂直平分线与x周的焦点的交点为T,求直线BT的斜率k
椭圆X^2/25+Y^2/9=1上不同三点A(x1,y1),B(4,9/5)C(x2,y2)与焦点F(4,0)的距离成等差数列
(1) 求证X1+x2=8
(2) 若线段AC的垂直平分线与x周的焦点的交点为T,求直线BT的斜率k
椭圆X^2/25+Y^2/9=1上不同三点A(x1,y1),B(4,9/5)C(x2,y2)与焦点F(4,0)的距离成等差数列(1) 求证X1+x2=8(2) 若线段AC的垂直平分线与x周的焦点的交点为T,求直线BT的斜率k
到右焦点F距离成等差数列等价于到右准线距离成等差数列,右准线方程为x = 25/4
也就是说 2(25/4-4) = (25/4-x1)+(25/4-x2) 所以 x1+x2 =8
设 AC中点M ,其坐标(x0,y0) ;所以 M(4,y0); AC斜率为 (x-4)*k1 = y-y0
其中垂线方程为 (x-4) = -k1*(y-y0),T:(4+k1y0,0).1#
现在用点差法探求k1与y0 的关系
9x1^2 + 25y1^2 = 225
9x2^2 + 25y2^2 =225 相减 9(x1-x2)(x1+x2) = -25(y1+y2)(y1-y2)
也就是说 36 = 9x0 = -25y0*k1; y0*k1 =-36/25 带回1#
可知T:(64/25,0)
因此BT斜率k = (9/5-0)/(4-64/25) = (9/5)/(36/25)=5/4
昨天我回答过一个 - -
看图