(1) P/2=1 ==>P=2,
y²=2Px=4x
(2) A(x1,y1),B(x₂,y₂),M(-1,y3)
已知F(1,0)是线段MA的中点,
(-1+x1)/2=1==>x1=3,y1=±2√3
当A(3,2√3)时,及中点:F(1,0)==>直线AB:y=√3(x-1)
当A(3,-2√3)时,及中点:F(1,0) ==>直线AB:y=-√3(x-1)
所以直线AB方程:y=√3(x-1)或者y=-√3(x-1)
(1) P/2=1 ==>P=2,
y²=2Px=4x
(2) A(x1,y1),B(x₂,y₂),M(-1,y3)
已知F(1,0)是线段MA的中点,
(-1+x1)/2=1==>x1=3,y1=±2√3
当A(3,2√3)时,及中点:F(1,0)==>直线AB:y=√3(x-1)
当A(3,-2√3)时,及中点:F(1,0) ==>直线AB:y=-√3(x-1)
所以直线AB方程:y=√3(x-1)或者y=-√3(x-1)