高中解几抛物线题
三角形ABC内接于抛物线y^2=16x,其中A(1,4),且三角形ABC的重心为抛物线的焦点,求直线BC的方程。...
三角形ABC内接于抛物线y^2=16x,其中A(1,4),且三角形ABC的重心为抛物线的焦点,求直线BC的方程。
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y^2 = 2*8x, 焦点F(4, 0)
AF的桐竖方程: (y - 4)/(x - 1) = (0-4)/(4 -1) = -4/3
4x + 3y -16 = 0
BC的中点M在直线AM上, M(m, (16-4m)/3)
|AF| = √[(4-1)² + (0-4)²] = 5
|AF| : |FM| = 2:1
|FM| = 5/2
|FM|² = 25/4 = (m - 4)² + [(16-4m)/3 - 0]² = (25/9)*(m-4)²
(m-4)² = 9/4
m-4 = ±3/2
m = 11/2, M(11/2, -2)
或嫌轮吵m = 5/2, M(5/2, 2) (与A都在轴上方,舍去)
B(b²/16, b)
C(c²/16, c)
BC的中点M((b²+c²)/32, (b+c)/2)
(b²+c²)/32 = 11/2
(b+c)/2 = -2
c = -2 ± √21
b = 2 ± √21
二者可芹侍互换, 这里取C(另一值为B):
C((11-√21)/2, -2 + √21)
CM(即BC的方程): (y + 2)/(x - 11/2) = (-2 + 2√21 +2)/[11-√21)/2 -11/2] = -4
4x + y -20 = 0
AF的桐竖方程: (y - 4)/(x - 1) = (0-4)/(4 -1) = -4/3
4x + 3y -16 = 0
BC的中点M在直线AM上, M(m, (16-4m)/3)
|AF| = √[(4-1)² + (0-4)²] = 5
|AF| : |FM| = 2:1
|FM| = 5/2
|FM|² = 25/4 = (m - 4)² + [(16-4m)/3 - 0]² = (25/9)*(m-4)²
(m-4)² = 9/4
m-4 = ±3/2
m = 11/2, M(11/2, -2)
或嫌轮吵m = 5/2, M(5/2, 2) (与A都在轴上方,舍去)
B(b²/16, b)
C(c²/16, c)
BC的中点M((b²+c²)/32, (b+c)/2)
(b²+c²)/32 = 11/2
(b+c)/2 = -2
c = -2 ± √21
b = 2 ± √21
二者可芹侍互换, 这里取C(另一值为B):
C((11-√21)/2, -2 + √21)
CM(即BC的方程): (y + 2)/(x - 11/2) = (-2 + 2√21 +2)/[11-√21)/2 -11/2] = -4
4x + y -20 = 0
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