设o为坐标原点,Rt△AOB内接于抛物线y^2=16x,若斜边AB长为32,求两直角边的斜率

Question

设o为坐标原点,Rt△AOB 内接于抛物线y^2=16x,若斜边AB长为32,求两直角边的斜率

Answer 1

设A点的坐标为(x1,y1),B点的坐标为(x2,y2)
∵Rt△AOB 内接于抛物线y^2=16x,斜边AB
∴OA⊥OB,(y1/x1)*(y2/x2)=-1
∵y^2=16x
∴y1/x1=16/y1,y2/x2=16/y2
∴(16/y1)*(16/y2)=-1即y1y2=-256
∴(y2-y1)²-(y2+y1)²=-4y1y2=32²
∵斜边AB长为32
∴(x2-x1)²+(y2-y1)²=32²
∴(x2-x1)²+(y2+y1)²=0
∴x2-x1=0,y2+y1=0
结合y1y2=-256,得y1=16,y2=-16,x1=x2=16或y1=-16,y2=16,x1=x2=16
∴y1/x1=1,y2/x2=-1或y1/x1=-1,y2/x2=1
两直角边的斜率是1和-1.