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白痴maths問題
2009-12-03 2:20 am
Find the slope of the tangent to the circle x^2+y^2-2x+4y-20=0 at x=1 and y=3
回答 (3)
2009-12-03 2:59 am
✔ 最佳答案
x^2+y^2-2x+4y-20=0Differentiate both sides with respect to x,
d(x^2+y^2-2x+4y-20) / dx = 0
2x + 2y (dy/dx) - 2 + 4 (dy /dx) = 0
(2y+4)dy/dx = 2 - 2x
dy / dx = (1-x)/(y+2)
At point (1,3),
dy/dx = (1-1)/(3+2) = 0
So the slope of the tangent at point (1,3) is 0.
2009-12-03 2:38 am
x2+y2-2x+4y-20=0
1x+3y-2[(1+x)/2]+4[(3+y)/2]-20=0
x+3y-(1+x)+2(3+y)-20=0
x+3y-1-x+6+2y-20=0
5y=15
y=3
∴the slope of the tangent=0
1x+3y-2[(1+x)/2]+4[(3+y)/2]-20=0
x+3y-(1+x)+2(3+y)-20=0
x+3y-1-x+6+2y-20=0
5y=15
y=3
∴the slope of the tangent=0
收錄日期: 2021-04-23 20:39:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091202000051KK02905