amaths

Solution:

圖片參考：http://hk.geocities.com/stevieg_1023/m0.gif


圖片參考：http://hk.geocities.com/stevieg_1023/m1.gif

a)
as m is slope of L with also pass through (0,0) so
m = (y1-0)/(x1-0) = y1/x1 and m = (y2-0)/(x2-0) = y2/x2

as H, K on C, so we have
x1^2 + y1^2 - 6x1 - 4y1 + 9 = 0 and x2^2 + y2^2 - 6x2 - 4y2 + 9 = 0
(1+(y1/x1)^2)*x1^2 - (6 - 4y1/x1)*x1 + 9 = 0 and
(1+(y2/x2)^2)*x2^2 - (6 - 4y2/x2)*x2 + 9 = 0
hence
(1+m^2)*x1^2 - (6 - 4m)*x1 + 9 = 0 and
(1+m^2)*x2^2 - (6 - 4m)*x2 + 9 = 0

ie. x1 and x2 are roots of equation (1+m^2)x^2-(6+4m)x+9=0

b)
sum of roots = -b/a
product of roots = c/a
i.e. x1+x2 = (6+4m)/(1+m^2) and x1*x2 = 9/(1+m^2)

(x1-x2)^2 = x1^2 - 2x1*x2 + x2^2 = x1^2 + 2x1*x2 + x2^2 - 4x1*x2
= (x1+x2)^2 - 4x1*x2 = ((6+4m)/(1+m^2))^2 - 4(9/(1+m^2))
= (36+48m+16m^2)/(1+m^2)^2 - (36+36m^2)/(1+m^2)^2
= (48m-20m^2)/(1+m^2)^2

c)
as m is slope of L, so m = (y1-y2)/(x1-x2)
HK^2 = (x1-x2)^2 + (y1-y2)^2 = (x1-x2)^2 *(1+(y1-y2)^2/(x1-x2)^2)
= (x1-x2)^2 *(1+((y1-y2)/(x1-x2))^2) = (x1-x2)^2 *(1+m^2)
= (48m-20m^2)/(1+m^2)^2 *(1+m^2)
= (48m-20m^2)/(1+m^2)

d)
for tangent to circle, we have HK = 0
i.e. solve (48m-20m^2)/(1+m^2) = 0
hence 48m-20m^2 = 0
4m(12- 5m) = 0
m = 0 or 12/5
as the tangent pass through origin
hence the two tangents are (y - 0)/(x - 0) = 0 or 12/5
i.e. y = 0 or 5y = 12x
the result are y = 0 or 12x - 5y = 0

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