Question

已知平面上一定点c（4,0）和一定直线L：x=1,p为该平面上的一动点,作PQ⊥L,垂足为Q,且（向量PC+2向量PQ）

Answer 1

(1) P(x, y), C(4, 0), Q(1, y)

向量PC = (4-x, 0-y) = (4-x, -y)

向量PQ = (1-x, 0)

2向量PQ = (2-2x, 0)

向量PC+2向量PQ = (4 - x +2 -2x, -y + 0) = (6 - 3x, -y)

向量PC- 2向量PQ = (4 - x - 2 + 2x, -y - 0) = (2 + x, -y)

(6-3x)(2+x) + (-y)^2 = 0

3x^2 -y^2 = 12

(2) 3x^2 -y^2 = 12, y = kx +1

解得A( [k - √(39-12k^2)]/(3-k^2), [3 - √(39-12k^2)]/(3-k^2) ) (k^2 ≠ 3)

B([k + √(39-12k^2)]/(3-k^2), [3 + √(39-12k^2)]/(3-k^2) )

AB的中点为圆心E(k/(3-k^2), 3/(3-k^2))

AB为直径d, d^2 = { [k - √(39-12k^2)]/(3-k^2) - k + √(39-12k^2)]/(3-k^2) }^2 +

{ [3 - √(39-12k^2)]/(3-k^2) - [3 + √(39-12k^2)]/(3-k^2) }^2

= 4(1+k^2)(39-12k^2)/(3-k^3) (1)

= 4r^2

r^2 = (1+k^2)(39-12k^2)/(3-k^3)

ED为半径r, r^2 = [k/(3-k^2) - 0]^2 + [ 3/(3-k^2) +2]^2 = (4k^4-35k^2+81)/(3-k^2)^2 (2)

由(1)(2),

(1+k^2)(39-12k^2) = 4k^4-35k^2+81

8k^4 -31k^2 + 21 = 0

(8k^2 - 7)(k^2 -3) = 0

k^2 = 7/8 或 k^2 = 3 (舍去)

k = ±√(7/8) = ±(√14)/4