Question

过点P(0,4)作直线x^2+y^2=4的切线L,若L与抛物线(p>0)交于两点A、B,且OA垂直OB,求抛物线的方程

Answer 1

切线kx-y+4=0

圆心到切线距离等于半径

4/√(k²+1)=2

k²+1=4

y=±√3x+4

y²=2px

则和y=-√3x+4有交点

3x²-(8√3+2p)x+16=0

x1+x2=(8√3+2p)/3

x1x2=16/3

y=-√3x+4

y1y2=(-√3x1+4)(-√3x2+4)

=3x1x2-√3(x1+x2)+16

=16-√3(8√3+2p)/3+16

=32-(24+2√3p)/3

OA,OB斜率y1/x1,y2/x2

垂直

y1y2/x1x2=-1

x1x2+y1y2=0

16/3+32-(24+2√3p)/3=0

p=44√3/3

y²=88√3x/3