Let C be the circle x^2 + y^2 + 4x - 8y + 2 = 0
and L be the line 2x - y + 4 = 0. L cuts C at two points A (x1, y1) and B (x2, y2).
a) Find the quadratic equation in x with x1 and x2
as roots.
b) Express the length AB in terms of x1 + x2 and x1x2 only.
c) Without solving the quadratic equations in (a), find the length of AB.
Equations of circle
2011-01-19 2:03 am
回答 (2)
2011-01-19 2:16 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_07Jan181815.gif
2011-01-22 1:15 am
(a)
C: x^2 + y^2 + 4x - 8y + 2 = 0
L: 2x - y + 4 = 0
Substituting y = 2x + 4 into C, we obtain
x^2 + (2x + 4)^2 + 4x - 8(2x + 4) + 2 = 0
x^2 + (4x^2 + 16x + 16) + 4x - (16x + 32) + 2 = 0
5x^2 + 4x - 14 = 0 ... (*)
x1 + x2 = -4/5
(x1)(x2) = -14/5
Quadratic equation becomes:
x^2 - (-4/5)x - 14/5 = 0
x^2 + 4/5 x - 14/5 = 0
2.
AB^2 = (x1 - x2)^2 + (y1 - y2)^2
= (x1 - x2)^2 + [(2 x1 + 4) - (2 x2 + 4)]^2
= (x1 - x2)^2 + (2 x1 - 2 x2)^2
= (x1 - x2)^2 + 4(x1 - x2)^2
= 5 (x1 - x2)^2
= 5 [(x1)^2 - 2(x1)(x2) + (x2)^2]
= 5 {[(x1)^2 + 2(x1)(x2) + (x2)^2] - 4(x1)(x2)}
= 5 {[(x1) + (x2)]^2 - 4(x1)(x2)}
AB = V {5{[(x1) + (x2)]^2 - 4(x1)(x2)}}
3.
From (a) and (b),
AB = V {5[(-4/5)^2 - 4(-14/5)]}
= V [5(296/25)]
= V 296/5
= V (4*74/5)
= 2 V(74/5)
C: x^2 + y^2 + 4x - 8y + 2 = 0
L: 2x - y + 4 = 0
Substituting y = 2x + 4 into C, we obtain
x^2 + (2x + 4)^2 + 4x - 8(2x + 4) + 2 = 0
x^2 + (4x^2 + 16x + 16) + 4x - (16x + 32) + 2 = 0
5x^2 + 4x - 14 = 0 ... (*)
x1 + x2 = -4/5
(x1)(x2) = -14/5
Quadratic equation becomes:
x^2 - (-4/5)x - 14/5 = 0
x^2 + 4/5 x - 14/5 = 0
2.
AB^2 = (x1 - x2)^2 + (y1 - y2)^2
= (x1 - x2)^2 + [(2 x1 + 4) - (2 x2 + 4)]^2
= (x1 - x2)^2 + (2 x1 - 2 x2)^2
= (x1 - x2)^2 + 4(x1 - x2)^2
= 5 (x1 - x2)^2
= 5 [(x1)^2 - 2(x1)(x2) + (x2)^2]
= 5 {[(x1)^2 + 2(x1)(x2) + (x2)^2] - 4(x1)(x2)}
= 5 {[(x1) + (x2)]^2 - 4(x1)(x2)}
AB = V {5{[(x1) + (x2)]^2 - 4(x1)(x2)}}
3.
From (a) and (b),
AB = V {5[(-4/5)^2 - 4(-14/5)]}
= V [5(296/25)]
= V 296/5
= V (4*74/5)
= 2 V(74/5)
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