某圓的方程是 kx(2x + y) - 4y(5x - hy) = 2011，其中k、h 是常數。求k + h 的
值
A circle has equation kx(2x + y) - 4y(5x - hy) = 2011, where k and h are constants.
Find the value of k + h
數學問題.求數學高手解答(20分)
2012-01-06 5:36 am
回答 (2)
2012-01-06 5:45 am
✔ 最佳答案
kx(2x + y) - 4y(5x - hy) = 20112kx^2 + kxy - 20xy + 4hy^2 = 2011
for a equation of circle:
2k = 4h
k = 20
=> h = 10
thus, k+h = 30
2012-01-06 5:49 am
kx(2x + y) - 4y(5x - hy) = 2011
2kx^2 + kxy - 20xy + 4hy^2 = 2011
2kx^2+4hy^2+(k-20)xy-2011=0
comparing to "x^2+y^2+Dx+Ey+F=0" ,
2k=4h and k-20=0
k=2h and k=20
k=20, h=10
k+h=30
2012-01-05 21:50:27 補充：
comparing to general circle form "x^2+y^2+Dx+Ey+F=0"
2kx^2 + kxy - 20xy + 4hy^2 = 2011
2kx^2+4hy^2+(k-20)xy-2011=0
comparing to "x^2+y^2+Dx+Ey+F=0" ,
2k=4h and k-20=0
k=2h and k=20
k=20, h=10
k+h=30
2012-01-05 21:50:27 補充：
comparing to general circle form "x^2+y^2+Dx+Ey+F=0"
收錄日期: 2021-04-23 20:02:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120105000051KK01118