✔ 最佳答案
(a)
Let the equation of L1 be y = 2x + c, where c > 0
Sub. y = 2x + c into x² + y² - 6x - 2y - 15 = 0,
x² + (2x + c)² - 6x - 2(2x + c) - 15 = 0
5x² + 2(2c - 5)x + (c² - 2c - 15) = 0
Since L1 is a tangent of C1, [2(2c - 5)]² - 4(5)(c² - 2c - 15) = 0
(2c - 5)² - 5(c² - 2c - 15) = 0
c² + 10c - 100 = 0
c = - 5 + 5√5 (since c > 0)
Thus, the equation of L1 is y = 2x - 5 + 5√5.
(b)
Sub. c = - 5 + 5√5 into 5x² + 2(2c - 5)x + (c² - 2c - 15) = 0
5x² + 2[2(-5 + 5√5) - 5]x + [(-5 + 5√5)² - 2(-5 + 5√5) - 15] = 0
25 - 50a + 125 +10 -10a - 15
x² + (4√5 - 6)x + (29 - 12√5) = 0
x = 3 - 2√5
Thus, the intersection is (3 - 2√5, 2(3 - 2√5) + (- 5 + 5√5))
i.e. (3 - 2√5, 1 + √5)
(c)
the centre of C1 is (3, 1)
the equation of L2 is y - 1 = 2(x - 3)
i.e. 2x - y - 5 = 0
(d)
Let the equation of L3 is y = 2x + d,
By similar triangles, (c - d) : 2 = √5 : 1
c - d = 2√5
(-5 + 5√5) - d = 2√5
d = -5 + 3√5
Thus, the equation of L3 is y = 2x - 5 + 3√5
(e)
radius of C1 = √[3² + 1² - (-15)] = 5
the distance of L2 and L3 = 5 - 2 = 3
the mid-pt. is ( [3(3 - 2√5) + 2(3)] / (3 + 2), [3(1 + √5) + 2(1)] / (3 + 2) )
i.e. (3 - 1.2√5, 1 + 0.6√5)
(f) and (g)
Sub. x = 0 into x² + y² - 6x - 2y - 15 = 0
y = -3, 5, so, c2 = -3
Let the required distance be D
By similar triangles, D : [-3 - (-5)] = 1 : √5
D = 0.4√5
Thus, the required distance is 0.4√5
Let the point be (p, q)
q = 2p - 5 and (q + 3) / p = -1 / 2,
we have p = 4/5 , q = -17/5
thus, the point is (4/5, 17/5)
