# 查看完整版本 : 圆与球

JYSim98 2015-9-26 01:14 PM

## 圆与球

1.一動點p分別到一定點（3,0）及圓心為（-3,0）r為5單位的一圓的圓周等距。求此動點p的軌跡方程式。

ANS：44x^2-100y^2-275

top11 2015-9-27 04:21 AM

[quote]原帖由 [i]JYSim98[/i] 於 2015-9-26 01:14 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426573368&ptid=25132862][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
1.一動點p分別到一定點（3,0）及圓心為（-3,0）r為5單位的一圓的圓周等距。求此動點p的軌跡方程式。

ANS：44x^2-100y^2-275 [/quote]

P 是 (3, 0) 和 (a, b) 的中點，即
{x = (3 + a)/2
{y = (0 + b)/2

{a = 2x - 3
{b = 2y

(2x - 3 + 3)² + (2y)² = 5²
[color=#ff0000]4x² + 4y² - 25 = 0[/color] 這就是 P(x, y) 的軌跡。

[url=https://www.wolframalpha.com/input/?i=4x%C2%B2+%2B+4y%C2%B2+-+25+%3D+0%2C+%28x+%2B+3%29%C2%B2+%2B+y%C2%B2+%3D+5%C2%B2%2C+x%3D3%2Cy%3D0%2C+44x%5E2-100y%5E2-275%3D0]https://www.wolframalpha.com/input/?i=4x%C2%B2+%2B+4y%C2%B2+-+25+%3D+0%2C+%28x+%2B+3%29%C2%B2+%2B+y%C2%B2+%3D+5%C2%B2%2C+x%3D3%2Cy%3D0%2C+44x%5E2-100y%5E2-275%3D0[/url]

JYSim98 2015-9-27 09:49 AM

[url=http://m.discuss.com.hk][img=100,23]http://n2.hk/d/images/r10/mobile.jpg[/img][/url]

top11 2015-9-27 11:58 AM

[quote]原帖由 [i]JYSim98[/i] 於 2015-9-27 09:49 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426631552&ptid=25132862][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

[img]http://n2.hk/d/images/r10/mobile.jpg[/img] [/quote]
Kamsah???
:smile_o06::smile_o06:

JYSim98 2015-9-27 02:19 PM

## 回覆 4# 的帖子

top11 2015-9-27 08:31 PM

[quote]原帖由 [i]JYSim98[/i] 於 2015-9-27 02:19 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426645821&ptid=25132862][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

mathtsang 2015-9-29 12:40 PM

I think the original answer is correct and top11's answer is wrong !
Since the question is to find the locus of a variable point P which is equi-distance to a given circle C and a given point G.
While top11's answer is given a variable point T on the given circle C and he find the locus of a variable point P which is equi-distance to the point T and another given point G.

top11 2015-9-29 10:17 PM

[quote]原帖由 [i]mathtsang[/i] 於 2015-9-29 12:40 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426774971&ptid=25132862][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
I think the original answer is correct and top11's answer is wrong !
Since the question is to find the locus of a variable point P which is equi-distance to a given circle C and a given point G.
Whi ... [/quote][color=#ff0000][/color]
[color=#ff0000]對，是我弄錯了，原答案正確，請看以下：[/color]

[color=#0000ff]√[(x + 3)² + y²] - 5 = √[(x - 3)² + y²]
(x + 3)² + y² - 10√[(x + 3)² + y²] + 25 = (x - 3)² + y²
(x + 3)² - (x - 3)² = 10√[(x + 3)² + y²] - 25
12x = 10√[(x + 3)² + y²] - 25
12x + 25 = 10√[(x + 3)² + y²]
144x² + 600x + 625 = 100[(x + 3)² + y²]
144x² + 600x + 625 = 100(x² + 6x + 9 + y²)
144x² + 600x + 625 = 100x² + 600x + 900 + 100y²
44x² + 625 = 900 + 100y²
44x² - 100y² - 275 = 0[/color]
[color=#ff0000]原答案正確。[/color]

[color=#008000]謝謝 mathtsang 指出我誤會了題目的地方！[/color]
:smile_o12: