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笔者在利用几何画板研究有心圆锥曲线的切线时发现一个简洁有趣的性质,现介绍如下:命题1自圆C_1:x~2+y~2=a~2+b~2上任一点P向椭圆C_2:x~2/a~2+y~2/b~2=1(a,b>0)引两条切线,则这两条切线互相垂直.证明:设P点的坐标为(x_0,y_0),自这一点向椭圆C_2引的两切线分别为l_1和l_2.(1)当切线的斜率存在且不为0时,设过P的切线方程为y-y_0=k(x-x_0),由y-y_0=k(x-x_0),x~2/a~2+y~2/b~2=1得(b~2+k~2a~2)x~2+
The author has found a concise and interesting character when using the geometrical drawing board to study the tangent of the conic curved concentric curve. Now we introduce the following: Proposition 1 From any point C 1: x ~ 2 + y ~ 2 = a ~ 2 + b ~ 2 P to ellipse C_2 The two tangent lines are perpendicular to each other when x ~ 2 / a ~ 2 + y ~ 2 / b ~ 2 = 1 (a, b> 0) leads to the fact that the coordinates of point P are (x_0, y_0 ), And the tangent lines leading to the ellipse C_2 from this point are respectively l_1 and l_2. (1) When the slope of the tangent exists and is not 0, the tangent equation of the set P is y-y_0 = k (x-x_0) (B ~ 2 + k ~ 2a ~ 2) x ~ 2 + 2 from y ~ y_0 = k (x-x_0) and x ~ 2 / a ~ 2 + y ~ 2 / b ~ 2 =