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  • 20171104 Sa Oct 消参


    2017-11-04 Sa

    $ P(-3, 0) $ 在圆C $ (x-3)^2 + y^2 = 8^2 $ 内,动圆M与圆相切且过P点,求M点轨迹。


    设切点 $ A(a, b) $,圆心 $M(x, y)$,则有 $R_M = MA = MP = R_C-MC$:

    $$ \left{ \begin{aligned}
    (a-3)^2 + b^2 &= 8^2 \
    \sqrt{(x-a)2+(y-b)2} &= \sqrt{(x+3)2+y2} \
    &= 8-\sqrt{(x-3)2+y2}
    \end{aligned} \right. $$

    Maxima:

    solve([(a-3)^2 + b^2 = 8^2, sqrt((x-a)^2+(y-b)^2) = sqrt((x+3)^2+y^2), sqrt((x+3)^2+y^2) = 8-sqrt((x-3)^2+y^2)],[x]);
    solve([(a-3)^2 + b^2 = 8^2], [a])
    
    (%i4) solve([(a-3)^2 + b^2 = 8^2], [a]);
                                        2                  2
    (%o4)           [a = 3 - sqrt(64 - b ), a = sqrt(64 - b ) + 3]
    
    solve([sqrt((x+3)^2+y^2) = 8-sqrt((x-3)^2+y^2)], [x])
    

    解不出来……


    UPD 2017-11-10 Fr 10:39PM

    周一在学校的时候想了一下,由 $ (a-3)^2 + b^2 = 8^2 $ 可知$a$ $b$关系,由 $ \sqrt{(x+3)2+y2} = 8-\sqrt{(x-3)2+y2} $ 可知 $x$ $y$ 关系,这样就只剩下两个未知数,然后再带入最后一个方程就行了。

    I simplfied in hand but the last step involved a polynomial with too much terms. So I decided to go to Maxima at weekend.

    Maxima:

    Maxima 5.25.0 http://maxima.sourceforge.net
    using Lisp Clozure Common Lisp Version 1.7-r14925M  (WindowsX8632)
    Distributed under the GNU Public License. See the file COPYING.
    Dedicated to the memory of William Schelter.
    The function bug_report() provides bug reporting information.
    (%i1) solve([(a-3)^2 + b^2 &= 8^2], [a])
    
    Maxima 5.25.0 http://maxima.sourceforge.net
    using Lisp Clozure Common Lisp Version 1.7-r14925M  (WindowsX8632)
    Distributed under the GNU Public License. See the file COPYING.
    Dedicated to the memory of William Schelter.
    The function bug_report() provides bug reporting information.
    (%i1) solve([(a-3)^2 + b^2 = 8^2], [a])
    ;
                                        2                  2
    (%o1)           [a = 3 - sqrt(64 - b ), a = sqrt(64 - b ) + 3]
    (%i2) a
    ;
    (%o2)                                  a
    (%i3) a;
    (%o3)                                  a
    (%i4) solve([sqrt((x-a)^2+(y-b)^2) = 8-sqrt((x-3)^2+y^2)], [y])
    ;
                 2            2            2    2              2    2
    (%o4) [sqrt(y  - 2 b y + x  - 2 a x + b  + a ) = 8 - sqrt(y  + x  - 6 x + 9)]
    (%i5) solve([sqrt((x-a)^2+(y-b)^2) = 8-sqrt((x-3)^2+y^2)], [y]);
                 2            2            2    2              2    2
    (%o5) [sqrt(y  - 2 b y + x  - 2 a x + b  + a ) = 8 - sqrt(y  + x  - 6 x + 9)]
    (%i6) solve([(x-a)^2+(y-b)^2 = 64 + (x-3)^2+y^2 - 16*sqrt((x-3)^2+y^2)], [y])
    ;
                           2    2                             2    2
                  16 sqrt(y  + x  - 6 x + 9) - 2 a x + 6 x + b  + a  - 73
    (%o6)    [y = -------------------------------------------------------]
                                            2 b
    (%i7) 
    

    Seems I don't know how to use Maxima to solve equation correctly...

    I gave it the wrong equation in %i4.. 'a' should not be involved. Try again:

    Maxima 5.25.0 http://maxima.sourceforge.net
    using Lisp Clozure Common Lisp Version 1.7-r14925M  (WindowsX8632)
    Distributed under the GNU Public License. See the file COPYING.
    Dedicated to the memory of William Schelter.
    The function bug_report() provides bug reporting information.
    (%i1) solve([sqrt((x+3)^2+y^2) = 8-sqrt((x-3)^2+y^2)], [y]);
                       2    2                        2    2
    (%o1)       [sqrt(y  + x  + 6 x + 9) = 8 - sqrt(y  + x  - 6 x + 9)]
    (%i2) solve([(x+3)^2+y^2 = 64 + (x-3)^2+y^2 - 16*sqrt((x-3)^2+y^2)], [y]);
    Is  3 x - 16  positive, negative, or zero?
    
    negative
    ;
                                         2                          2
                      sqrt(7) sqrt(16 - x )      sqrt(7) sqrt(16 - x )
    (%o2)      [y = - ---------------------, y = ---------------------]
                                4                          4
    (%i3) 
    

    Well.. let's do it step by step.

    $$ \left{ \begin{aligned}
    (a-3)^2 + b^2 &= 8^2 \
    \sqrt{(x-a)2+(y-b)2} &= \sqrt{(x+3)2+y2} \
    8-\sqrt{(x-3)2+y2} &= \sqrt{(x+3)2+y2}
    \end{aligned} \right. $$

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  • 原文地址:https://www.cnblogs.com/yanhuihang/p/7783830.html
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