圆锥曲线的切线方程和切点弦方程的证明

第一种方法：判别式法。点P(x0,y0)在椭圆上，可以设P点的切线方程为y−y0=(x−x0)代入椭圆方程，利用判别式等于0，解出k。第二种方法：隐函数求导x2a2+y2b2=12xa2+2yy′b2=0xa2=−yy′b2y′=−b2xa2y点p(x0,y0)处切线斜率为−b2x0a2y0切线方程为y−y0=−b2x0a2y0(x−x0)a2y0y−a2y02=−b2x0(x−x0)a2y0y+b2x0x=b2x02+a2y02因为点p(x0,y0)在椭圆上，x02a2+y02b2=1所以b2x02+a2y02=a2b2所以a2y0y+b2x0x=b2x02+a2y02=a2b2x0xa2+y0yb2=1同理双曲线点P(x0,y0)处的切线方程为x0xa2−y0yb2=1抛物线点P(x0,y0)处的切线方程为y0y=p(x+x0)下面证明椭圆外一点P(x0,y0)向椭圆做两条切线PA和PB，切点为A(x1,y1)B(x2,y2)切点弦所在的直线方程为x0xa2+y0yb2=1切线PA的方程和切线PB的方程分别为x1xa2+y1yb2=1x2xa2+y2yb2=1两式相减得x(x1−x2)a2=−y(y1−y2)a2−b2xa2y=(y1−y2)(x1−x2)因为点P(x0,y0)是两条切线的交点所以−b2x0a2y0=(y1−y2)(x1−x2)所以直线AB的斜率k=−b2x0a2y0直线AB的点斜式方程为y−y1=−b2x0a2y0(x−x1)y−y2=−b2x0a2y0(x−x2)两式相加得2y−(y1+y2)=−b2x0a2y0[2x−(x1+x2)]切线PA的方程和切线PB的方程分别为x1xa2+y1yb2=1x2xa2+y2yb2=1两式相加得x(x1+x2)a2=y(y1+y2)a2+2y1+y2=b2x0(x1+x2)−2a2b2a2y0把y1+y2代入2y−(y1+y2)=−b2x0a2y0[2x−(x1+x2)]化简后得x0xa2+y0yb2=1同理过双曲线外一点P(x0,y0)向双曲线做两条切线PA和PB，切点为A(x1,y1)B(x2,y2)切点弦所在的直线方程为x0xa2−y0yb2=1同理过抛物线外一点P(x0,y0)向抛物线做两条切线PA和PB，切点为A(x1,y1)B(x2,y2)切点弦所在的直线方程为y0y=p(x+x0)