(a) Given teo points A(-1,2) and B(3,0), find the equation of the locus of a point P such that AP=kBP, where k>0.
The equation of the locus point P is
(1-k^2)x^2 + (1-k^2)y^2 + (2+6k^2)x-4y+5-9k^2=0
(b) State the value of k for which the locus in (a) is a straight line. (Ans: 1)
我懂得做 Q(a), 但是我不懂得做Q(b). 只要教我如何找出the value of k 就可以了。麻煩了!
S.5 math's Q
2013-05-11 5:04 am
回答 (1)
2013-05-11 6:51 am
✔ 最佳答案
很簡單- -因為直線方程不會有2次項
只要x^2同y^2的coefficient=0就得了
1-k^2=0
k^2=1
k=1 or -1(rej, k>0)
收錄日期: 2021-04-13 19:27:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130510000051KK00222