F.5 MATHS

(a)
L1: 2x - y + q = 0 …… (1)
C: 5x² + 5y² = 16 …… (2)

(1): y = 2x + q …… (3)

Put (3) into (2):
5x² + 5(2x + q)² = 16
5x² + 20x² + 20qx + 5q² = 16
25x² + 20qx + (5q² - 16) = 0 …… (4)

Since L1 is the tangent of C, then determinant Δ = 0
(20q)² - 4(25)(5q² - 16) = 0
400q² - 500q² + 1600 = 0
100q² = 1600
q² = 16
q = 4 or q = -4 (rejected)

L2: 2x + ry - 20 = 0 …… (5)
C: 5x² + 5y² = 16 …… (2)

(5): x = (20 - ry)/2 …… (6)

Put (6) into (2):
5[(20 - ry)/2]² + 5y² = 16
2000 - 200ry + 5r²y² + 20y² = 64
(5r² + 20)y² - 200ry + 1936 = 0 …… (7)

Since L2 is the tangent of C, then determinant Δ = 0
(200r)² - 4(5r² + 20)(1936) = 0
40000r² - 38720r² - 154880 = 0
1280r² = 154880
r² = 121
r = 11 or r = -11 (rejected)


(b)
L1 : 2x - y + 4 = 0 …… (1)
C: 5x² + 5y² = 16 …… (2)

(1): y = 2x + 4 …… (3)

Put (3) into (2):
5x² + 5(2x + 4)² = 16
5x² + 20x² + 80x + 6400 = 16
25x² + 80x + 64 = 0
(5x + 8)² = 0
x = -8/5 (double roots)

Put x = -8/5 into (3):
y = 2(-8/5) + 4
y = 4/5

Hence, the intersection between L1 and C is (-8/5, 4/5).

L2: 2x + 11y - 20 = 0 …… (5)
C: 5x² + 5y² = 16 …… (2)

(5): x = (20 - 11y)/2 …… (6)

Put (6) into (2):
5[(20 - 11y)/2]² + 5y² = 16
2000 - 2200y + 605y² + 20y² = 64
625y² - 2200y + 1936 = 0
(25y - 44) = 0
y = 44/25

Put y = 44/25 into (6):
x = [20 - 11(44/25)]/2
x = 8/25

Hence, the intersection between L2 and C is (8/25, 44/25).