中4方程兩條

2009-05-16 10:54 pm
1)解方程X^2-2Y^2+2Y=3X+4Y=4

2)若直線Y=MX+C與曲線Y=X^2-5X+10相切於點(4,6),求C和M的值

回答 (1)

2009-05-16 11:27 pm
✔ 最佳答案
1)
x2 - 2y2 + 2y = 4 ...... (1)
3x + 4y = 4 ...... (2)

(2):
x = (4 - 4y)/3 ...... (3)

[(4 - 4y)/3]2 - 2y2 + 2y = 4
9{[(4 - 4y)/3]2 - 2y2 + 2y} = 9 x 4
(4 - 4y)2 - 18y2 + 18y = 36
16 - 32y + 16y2 - 18y2 + 18y = 36
2y2 +14y + 20 = 0
y2 + 7y + 10 = 0
(y + 2)(y + 5) = 0
y = -2 or y = -5

When y = -2:
(3):
x = [4 - 4(-2)]/3
x = 4

When y = -5:
(3):
x = [4 - 4(-5)]/3
x = 8

Ans: x = 4, y = -2 or x = 8, y = -5


2)
y = x2 - 5x + 10 ...... (1)
y = mx + c ...... (2)

(2)→(1):
mx + c = x2 - 5x + 10
x2 - (m + 5)x2 + (10 - c) = 0
由於只有一個切點,Δ = 0
[-(m + 5)]2 - 4(1)(10 - c) = 0
m2 + 10m + 25 - 40 + 4c = 0
m2 + 10m + 4c - 15 = 0 ...... (3)

(4, 6) 在切線 y = mx + c 上:
6 = m(4) + c
c = 6 - 4m ...... (4)

(4)→(3)
m2 + 10m + 4(6 - 4m) - 15 = 0
m2 + 10m + 24 - 16m - 15 = 0
m2 - 6m + 9 = 0
(m - 3)2 = 0
m = 3 (重根)

(4):
c = 6 - 4(3)
c = -6

答:m = 3, c = -6
=


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