Find the equation of the tangent to the circle x^2 + y^2 -4x+4y+6=0
at the point (3,-1) on the circle.
我想問除左3x+(-1)y+ 1/2 (-4)(x+3) +1/2 (4)(y-1)+6=0呢個之外
將佢整成
eq. of tangent set 成: y+1= m(x-3)
then 搵centre , radius ,再 put 入去
呢個方法for tangent to a circle at a point on the circle ok ma??
我用呢個方法計到
x+y-2=0 呢個用第一個方法同上面個 都計到 ,and is the ans
7x-y-22=0 呢個係上面計到 ,但ans 冇呢個
呢個點解要rej. ??? plz explain~
then, 可唔可以講解下係睇d咩而去rej. ??
thz~!
A.maths 高手請答~
2007-05-16 7:41 am
回答 (2)
2007-05-16 8:22 am
✔ 最佳答案
其實用tangent set係可以計到重點係點搵m
簡單D黎講
而家既info有circle上的一點and centre (2,-2) 既co-ordinates
你可以搵到D條線既slope (即 +1)
因為tangent at a point perpendicular to radius
所以tangent 的slope is -1
可以出到 x+y+2=0
至於你的7x-y-22=0其實唔係好知你點計到, 任何情況下tangent at a point都係得一條線
如果你用radius 去搵,應該只會有double root
proof:
y=7x-22
x^2+(7x-22)^2-4x+4(7x-22)+6=0
50x^2+332x+462=0
discriminant = 332^2 - 4 * 50 * 462 =17824 > 0
2 real roots and therefore not a tangent.
2007-05-16 8:13 am
你又無講點用centre , radius ,再 put 入去... 唔知你過程出錯乜..
不過你要計都有得計
點同直線距離, normal form 你有無學過
centre(2,-2) , radius = sqrt(2)
tangent mx - y - 3m - 1 = 0
點同直線距離 = abs [ 2m - (-2) - 3m - 1 ] / sqrt[ m^2 + (-1)^2 ] = sqrt(2)
m = -1
得一個答案, 即係用第一個方法計果個...
可能你計錯數
不過你要計都有得計
點同直線距離, normal form 你有無學過
centre(2,-2) , radius = sqrt(2)
tangent mx - y - 3m - 1 = 0
點同直線距離 = abs [ 2m - (-2) - 3m - 1 ] / sqrt[ m^2 + (-1)^2 ] = sqrt(2)
m = -1
得一個答案, 即係用第一個方法計果個...
可能你計錯數
收錄日期: 2021-04-13 00:29:04
原文連結 [永久失效]:
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