抛物线y²=2px焦点为F,O为坐标原点,M为抛物线上一点,且MF=4OF...答:F(0.5p,0)|OF|=0.5|p| |MF|=4|OF|=2|p| (x-0.5p)^2+y^2=4p^2 (x-0.5p)^2+2px=4p^2 (x+0.5p)^2=(2p)^2 xM=1.5p,-2.5p |yM|=|√3|p|,√5|p| s1=(1/2)*0.5p*√3p=√3,p=±2,y^2=±4x s2=(1/2)*0.5p*√5p=√3,p=-0.8√15,0.8√...
过P点做抛物线的切线,求三角形PAB面积最小值。答:抛物线焦点为F(0,p/2)=F(0,1)设切点为A(x1,y1),B(x2,y2),设焦点弦方程为y=kx+1,代入抛物线得 x^2=4y=4kx+4, 即x^2-4kx-4=0 由韦达定理可得 x1+x2=4k, x1x2=-4 ∴|x1-x2|=√[(x1+x2)^2-4x1x2]设焦点弦中点为M(m,n),则 m=(x1+x2)/2=2k, n=km+1=...