Equation of Tangent

[匿名]

2013-10-12 7:52 am

picture of Q 1 & Q 2 : http://upload.lsforum.net/users/public/p2369711-10j64.png

1) Two tangents L(1) and L(2) are drawn from the origin to the circle
x^+y^-10x-6y+9=0 If θ is the angle between L(1) and L(2), find
a) the radius of the circle and the length of the tangents from the origin to the
circle.
b) the value of tanθ without using a calculator

I know how to do (a) but can;t do (b), and I want to ask a Q:
如果我想計slope of tangent, 咁 radius=5, centre(5,3)
let eq of tangent: y=mx+c <---由於pass through origin, 所以eq of tangent: y=mx
之後用perpendicular distance of centre to tangent=radius計m, 但我只計到一個
slope,就是-8/15, why?????

2) A straight line L with slope m passes through the origin and cuts the circle C :
x^+y^-6x-4y +9=0 at two points H and K
a) http://upload.lsforum.net/users/public/w4763611-102w64.png
Picture : http://upload.lsforum.net/users/public/p2369711-10j64.png

回答 (1)

用戶4366

2013-10-12 9:37 am

✔ 最佳答案

請參閱下圖，不用次次計斜率，用三角學都得！

圖片參考：http://imgcld.yimg.com/8/n/HA00388954/o/20131012013507.jpg

2013-10-12 01:55:58 補充：
x^2 + y^2 - 6x - 4y + 9 = 0 ... (C)
y=mx
sub y=mx into (C) ==>
x^2 + m^2x^2 - 6x - 4mx + 9 = 0
(1+m^2)x^2 - (6+4m)x + 9 = 0

roots of this equation is where y=mx cuts (C), of course they are X1 and X2.

2013-10-12 02:00:24 補充：
Sum of roots X1+X2=(6+4m)/(1+m^2)
product of root (X1)(X2) = 9/(1+m^2)

(X1 - X2)^2 = (X1 + X2)^2 - (4)(X1)(X2)

(X1 - X2)^2 = (6+4m)^2/(1+m^2)^2 - 36/(1+m^2)

= (36+48m+16m^2-36-36m^2)/(1+m^2)^2

= (48m - 20m^2)/(1+m^2)^2

2013-10-12 02:03:46 補充：
sub x=y/m into (C) would give (Y1 - Y2)^2 = (48m^3 - 20m^4)/(1+m^2)^2

(HK)^2 = (X1 - X2)^2 + (Y1 - Y2)^2

Substitution would give the answer you want.

2013-10-12 02:07:23 補充：
If X1=X2 and Y1=Y2, H coincides with K and the line is a tangent.

HK = 0 ; 48m - 20m^2 = 0

(48-20m)m=0

m=0 or m=48/20=12/5 ;

then y=mx

2013-10-12 02:26:19 補充：
做這些數一定唔可以怕繁，俾心機啦！小心做運算呀！好易出錯。

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