一道解析几何问题已知抛物线y^2=2px(p>0)(1)过抛物线的焦点为2的直线l交抛物线于A,B两点,若|AB|=2,
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一道解析几何问题
已知抛物线y^2=2px(p>0)
(1)过抛物线的焦点为2的直线l交抛物线于A,B两点,若|AB|=2,求p的值;
(2)过点M(2p,0)作任何直线l交抛物线于P,Q两点,求证:OP⊥OQ.
已知抛物线y^2=2px(p>0)
(1)过抛物线的焦点为2的直线l交抛物线于A,B两点,若|AB|=2,求p的值;
(2)过点M(2p,0)作任何直线l交抛物线于P,Q两点,求证:OP⊥OQ.
![一道解析几何问题已知抛物线y^2=2px(p>0)(1)过抛物线的焦点为2的直线l交抛物线于A,B两点,若|AB|=2,](/uploads/image/z/17040128-32-8.jpg?t=%E4%B8%80%E9%81%93%E8%A7%A3%E6%9E%90%E5%87%A0%E4%BD%95%E9%97%AE%E9%A2%98%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%5E2%3D2px%28p%3E0%29%281%29%E8%BF%87%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E7%84%A6%E7%82%B9%E4%B8%BA2%E7%9A%84%E7%9B%B4%E7%BA%BFl%E4%BA%A4%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9%2C%E8%8B%A5%7CAB%7C%3D2%2C)
N(-1,0)
直线L:x=ty+1,与抛物线y2=4x联立后得
y^2-4ty-4=0,
y1+y2=4t,y1y2=-4
(1)kNA+kNB=y1/(y1^2/4 + 1) +y2/(y2^2/4 + 1)
=[1/4y1y2^2+1/4y1^2y2+y1+y2]/(y1^2/4 + 1)(y2^2/4 + 1)
=(y1y2/4 +1)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1)
=(-1+1)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1) =0
(2)S=1/2*|AB|*d
d=|-2|/√(1+t^2)=2/√(1+t^2)
|AB|=√(1+t^2)|y1-y2|=√(1+t^2)*√[(y1+y2)^2-4y1y2]
=√(1+t^2)*√16(1+t^2)
=4(1+t^2)
S=1/2*|AB|*d
=1/2*4(1+t^2)*2/√(1+t^2)
=4√(1+t^2)
当t=0,Smin=4
(3)若M(m,0)时,(1)仍成立
直线L:x=ty+m,与抛物线y2=4x联立后得
y^2-4ty-4m=0,
y1+y2=4t,y1y2=-4m
(1)kNA+kNB=y1/(y1^2/4 + m) +y2/(y2^2/4 + m)
=[1/4y1y2^2+1/4y1^2y2+my1+my2]/(y1^2/4 + m)(y2^2/4 + m)
=(y1y2/4 +m)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1)
=(-m+m)(y1+y2)/(y1^2/4 + m)(y2^2/4 + m) =0
(2)S=1/2*|AB|*d
d=|-2m|/√(1+t^2)=|2m|/√(1+t^2)
|AB|=√(1+t^2)|y1-y2|=√(1+t^2)*√[(y1+y2)^2-4y1y2]
=√(1+t^2)*√16(m+t^2)
S=1/2*|AB|*d
=1/2*√(1+t^2)*√16(m+t^2)*|2m|/√(1+t^2)
=|m|*√16(m+t^2)
=4√m^2(m+t^2)
令u=m^2(m+t^2),u'=2m^2*t=0,
当t>0,u'>0,当t<0,u'<0
t=0是极小值点,
当t=0,Smin=4√m^3=4m*√m
直线L:x=ty+1,与抛物线y2=4x联立后得
y^2-4ty-4=0,
y1+y2=4t,y1y2=-4
(1)kNA+kNB=y1/(y1^2/4 + 1) +y2/(y2^2/4 + 1)
=[1/4y1y2^2+1/4y1^2y2+y1+y2]/(y1^2/4 + 1)(y2^2/4 + 1)
=(y1y2/4 +1)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1)
=(-1+1)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1) =0
(2)S=1/2*|AB|*d
d=|-2|/√(1+t^2)=2/√(1+t^2)
|AB|=√(1+t^2)|y1-y2|=√(1+t^2)*√[(y1+y2)^2-4y1y2]
=√(1+t^2)*√16(1+t^2)
=4(1+t^2)
S=1/2*|AB|*d
=1/2*4(1+t^2)*2/√(1+t^2)
=4√(1+t^2)
当t=0,Smin=4
(3)若M(m,0)时,(1)仍成立
直线L:x=ty+m,与抛物线y2=4x联立后得
y^2-4ty-4m=0,
y1+y2=4t,y1y2=-4m
(1)kNA+kNB=y1/(y1^2/4 + m) +y2/(y2^2/4 + m)
=[1/4y1y2^2+1/4y1^2y2+my1+my2]/(y1^2/4 + m)(y2^2/4 + m)
=(y1y2/4 +m)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1)
=(-m+m)(y1+y2)/(y1^2/4 + m)(y2^2/4 + m) =0
(2)S=1/2*|AB|*d
d=|-2m|/√(1+t^2)=|2m|/√(1+t^2)
|AB|=√(1+t^2)|y1-y2|=√(1+t^2)*√[(y1+y2)^2-4y1y2]
=√(1+t^2)*√16(m+t^2)
S=1/2*|AB|*d
=1/2*√(1+t^2)*√16(m+t^2)*|2m|/√(1+t^2)
=|m|*√16(m+t^2)
=4√m^2(m+t^2)
令u=m^2(m+t^2),u'=2m^2*t=0,
当t>0,u'>0,当t<0,u'<0
t=0是极小值点,
当t=0,Smin=4√m^3=4m*√m
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