已知过点A的直线y=x 1叫抛物线与另一点E
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①存在斜率k,则直线方程:y=k(x-4)x=y/k+4y^2=4x联立y^2-4y/k-16=0y1^2+y2^2=(4/k)^2-(-16)*2>32②不存在斜率k,则x=4y^2=4xy1^2+
(1)证明:∵y=x24,∴y′=x2,∴kl=y′|x=x1=x12,∴l:y=x12(x−x1)+x124=x12x−x124,∴C(x12,0),设H(a,-1),∴D(a,0),∴TH:y=-
设l:x=my+1,与抛物线方程联立消x,可得y1*y2,y1+y2,再可得x1*x2.x1+x2,向量TA·向量TB=1用x1x2y1y2表示可得m,1/m即为斜率
设AB:y=k(x+2)设A(x1,y1),B(x2,y2),C(x3,y3),D(x4,y4)∴AM的方程是y=[y1/(x1-1)](x-1)设k0=y1/(x1-1)则AM:y=k0(x-1)与
(1)∵抛物线过点A(2,4)∴4a+2b+c=4……①∵x1²+x2²=13即(x1+x2)²-2x1x2=13又∵x1+x2=-b/a,x1x2=c/a(韦达定理)则
32联立y=k(x-4)y^2=4x=k^2*(x^2-8x+16)得x1+x2=8+4/(k^2)y1^2+y2^2=4(x1+x2)大于32或斜率不存在,得32
(1)抛物线焦点(0,1/4)所以设直线为y-1/4=kxy=kx+1/4带入抛物线kx+1/4=x^2x^2-kx-1/4=0根据韦达定理x1x2=-1/4/1=-1/4(2)AP=(x0-x1,y
y²=8xy1y2=16(y1)²=8x1(y2)²=8x2(y1y2)=64x1x2(16)²=64x1x2x1x2=4设A(0,b)直线方程y-0=k(x-
1:设y-2=k(x-1)因为平行,L的斜率=-1/2即k=-1/2答案:x+2y-5=02:因为垂直,所以k=2答案:2x-y=0
过点P(4,0)的直线的斜率显然不为0所以其方程可设为x=my+4,代入Y∧2=4X得y²-4my-16=0∴y1+y2=4m,y1y2=-16∴Y1∧2+Y2∧2=(y1+y2)²
A.4焦点(p/2,0)直线方程y=k(x-p/2)y^2=k^2x^2-k^2px+k^2p^2/4-2px=0k^2x^2-(k^2p+2p)x+k^2p^2/4=0x1x2=p^2/4(y1^2
易知直线l与x轴夹角为30°∴By=Ay=1,Bx=Ay•√(3)=√(3)A1y=Bx•√(3)+Ay=3+1=4,B1x=A1y•√(3)=4√(3)A2y=B
因为y1=kx1+by2=kx2+bky1
y=-2x+bk=-2,y随x的增大而减小∵x1
1、焦点(p/2,0)若垂直x轴,是x=p/2则y²=p²y1=-p,y2=py1y2=-p²若有斜率y=k(x-p/2)x=y/k+p/2所以y²=2py/k
设直线y=k(x-m)y/k=x-mx=y/k+mx=y/k+m代入y^2=2pxy^2=2p(y/k+m)y^2-(2p/k)y-2pm=0由根与系数的关系y1y2=-2pm又y1y2=-2m所以-
2(x1-x2)-3(y1-y2)=0(y1-y2)/(x1-x2)=2/3=k即直线斜率所以y-y1=2/3(x-x1)3y-3y1=2x-2x13y=2x-(2x1-3y1)=2x-4即y=2x/
y=3/x,y=kx+2(kx+2)x=3kx^2+2x-3=0x1+x2=-2/k,x1x2=-3/kX1*X1+X2*X2=10(x1+x2)^2-2x1x2=104/k^2+6/k=105k^2
由y=kx+b平行于y=-2x,∴k=-2.将A(-1,3)代入:3=2+b,∴b=1即y=-2x+1.
焦点(p/2,0)设直线AB:y=2√2(x-p/2)代入y²=2px得4x²-5px+p²=0x1+x2=5p/4|AB|=x1+x2+p=9p/4=9p=4即抛物线y