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.
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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.](/uploads/image/z/7270532-44-2.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%E6%B1%82p%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%BF%87%E7%82%B9M%282p%2C0%29%E4%BD%9C%E4%BB%BB%E4%BD%95%E7%9B%B4%E7%BA%BFl%E4%BA%A4%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%BA%8EP%2CQ%E4%B8%A4%E7%82%B9%2C%E6%B1%82%E8%AF%81%EF%BC%9AOP%E2%8A%A5OQ.)
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.
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.
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.
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
(1)焦点为(p/2,0),若知焦点坐标可求出p,
(2)当l⊥x轴时,把x=2p代入y^2=2px得y= ±2p,
P,Q坐标分别为(2p,2p),(2p,-2p)或(2p,-2p),(2p,2p).
|PM|=|QM|=|OM|,∠POM=∠QOM=45°,∠POQ=∠POM+∠QOM=90°,OP⊥OQ.
当l不垂直于x轴时,设l方程为y=k(x-2p),代入...
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(1)焦点为(p/2,0),若知焦点坐标可求出p,
(2)当l⊥x轴时,把x=2p代入y^2=2px得y= ±2p,
P,Q坐标分别为(2p,2p),(2p,-2p)或(2p,-2p),(2p,2p).
|PM|=|QM|=|OM|,∠POM=∠QOM=45°,∠POQ=∠POM+∠QOM=90°,OP⊥OQ.
当l不垂直于x轴时,设l方程为y=k(x-2p),代入y^2=2px得k²x²-(4k²+2)px+4k²p²=0,
设P,Q坐标分别为(x1,y1),(x2,y2).
则x1•x2=4k²p²/k²=4p²,(y1)²(y2)²=(y1•y2)²=2px1•2px2=(2p)²(x1•x2)=(2p)²•4p²=(4p²)²,
P,Q在x轴的上下两侧,0>y1•y2=-√[(4p²)²]=-4p²
直线OP,直线OQ的斜率分别为KOP,KOQ,
KOP=y1/x1,KOQ=y2/x2,KOP•KOQ=(y1/x1)•(y2/x2)=y1•y2/(x1•x2)=-4p²|/(4p²)=-1.
直线OP,直线OQ的斜率之积=-1,OP⊥OQ.
综上,OP⊥OQ.
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当年轻易的题居然已经不熟悉了55555555
你题目写错看,什么为2啊