难道这题数学题就没有人会了吗?抛物线C的方程为: y^2 = 2px (p>0),点P(1,2),A(X1,Y1),B(X2,Y2)均在抛物线上.求:当直线PA,PB的斜率存在且倾斜角互补时,求y1+y2的值及直线AB的斜率
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![难道这题数学题就没有人会了吗?抛物线C的方程为: y^2 = 2px (p>0),点P(1,2),A(X1,Y1),B(X2,Y2)均在抛物线上.求:当直线PA,PB的斜率存在且倾斜角互补时,求y1+y2的值及直线AB的斜率](/uploads/image/z/13390461-45-1.jpg?t=%E9%9A%BE%E9%81%93%E8%BF%99%E9%A2%98%E6%95%B0%E5%AD%A6%E9%A2%98%E5%B0%B1%E6%B2%A1%E6%9C%89%E4%BA%BA%E4%BC%9A%E4%BA%86%E5%90%97%3F%E6%8A%9B%E7%89%A9%E7%BA%BFC%E7%9A%84%E6%96%B9%E7%A8%8B%E4%B8%BA%3A+y%5E2+%3D+2px+%28p%3E0%29%2C%E7%82%B9P%281%2C2%29%2CA%28X1%2CY1%29%2CB%28X2%2CY2%29%E5%9D%87%E5%9C%A8%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8A.%E6%B1%82%3A%E5%BD%93%E7%9B%B4%E7%BA%BFPA%2CPB%E7%9A%84%E6%96%9C%E7%8E%87%E5%AD%98%E5%9C%A8%E4%B8%94%E5%80%BE%E6%96%9C%E8%A7%92%E4%BA%92%E8%A1%A5%E6%97%B6%2C%E6%B1%82y1%2By2%E7%9A%84%E5%80%BC%E5%8F%8A%E7%9B%B4%E7%BA%BFAB%E7%9A%84%E6%96%9C%E7%8E%87)
难道这题数学题就没有人会了吗?抛物线C的方程为: y^2 = 2px (p>0),点P(1,2),A(X1,Y1),B(X2,Y2)均在抛物线上.求:当直线PA,PB的斜率存在且倾斜角互补时,求y1+y2的值及直线AB的斜率
难道这题数学题就没有人会了吗?
抛物线C的方程为: y^2 = 2px (p>0),点P(1,2),A(X1,Y1),B(X2,Y2)均在抛物线上.
求:当直线PA,PB的斜率存在且倾斜角互补时,求y1+y2的值及直线AB的斜率
难道这题数学题就没有人会了吗?抛物线C的方程为: y^2 = 2px (p>0),点P(1,2),A(X1,Y1),B(X2,Y2)均在抛物线上.求:当直线PA,PB的斜率存在且倾斜角互补时,求y1+y2的值及直线AB的斜率
tan(a)=(2-y1)/(1-x1)
tan(b)=(2-y2)/(1-x2)
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)*tan(b))
PA与PB的倾斜角互补
所以0=tan(a)+tan(b)
即(2-y1)/(1-x1)+(2-y2)/(1-x2)=0
可得:(2-y1)(1-x2)+(2-y2)(1-x1)
=(2-y1)(1-1/4*y2^2)+(2-y2)(1-1/4*y1^2)
=1/4*(y1-2)(y2-2)(4+y1+y2)
=0
所以y1=2或y2=2或y1+y2=-4
因为PA与PB的斜率存在,所以y1=2或y2=2都舍去.
所以y1+y2=-4
(y2-y1)/(x2-x1)
=4*(y2-y1)/(y2^2-y1^2)
=4/(y2+y1)
=-1
法二:
A(x1,y1)B(x2,y2)在抛物线上,必然满足解析式Y^2=4X,
y1^2=4x1,x1=y1^2/4,
y2^2=4x2,x2=y2^2/4,
即:A(y1^2/4 ,y1) B(y2^2/4,y2)
kPA=(y1-2)/(x1-1)=(y1-2)/(y1^2/4-1)=4/(y1+2),
kPB=(y2-2)/(x2-1)=(y2-2)/(y2^2/4-1)=4/(y2+2),
PA与PB的斜率存在且倾斜角互补,说明kPA=-kPB,
即:4/(y1+2)=-4/(y2+2),
y1+2=-(y2+2),
y1+y2=-4;
kAB=(Y2-Y1)/(X2-X1)
=(y2-y1)/(y2^2/4-y1^2/4)
=4/(y1+y2)
=4/(-4)
=-1.