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Math Help - [SOLVED] Locus

  1. #1
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    Exclamation [SOLVED] Locus

    Hi guys

    A ladder that is 6 m long rests with one end on the horizontal ground and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axis respectively, find the locus of the midpoint of the ladder.

    For this question, I havn't been able to get far ...

    I let the pt the ladder is touching the horizontal ground (and thus the X-axis) be x AND the pt the ladder is touching the wall (the Y-axis) be y. Thus, the midpt is (x/2,y/2).

    I've tried to work it out the same method as the previous q.s in the txtbk e.g. saying this distance = that distance to get the locus.

    Please please help me!
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi guys

    A ladder that is 6 m long rests with one end on the horizontal ground and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axis respectively, find the locus of the midpoint of the ladder.

    For this question, I havn't been able to get far ...

    I let the pt the ladder is touching the horizontal ground (and thus the X-axis) be x AND the pt the ladder is touching the wall (the Y-axis) be y. Thus, the midpt is (x/2,y/2).

    I've tried to work it out the same method as the previous q.s in the txtbk e.g. saying this distance = that distance to get the locus.

    Please please help me!
    Let F denote the footpoint of the ladder. Then F(t, 0).

    then the height of the ladder against the wall is

    h = \sqrt{36 - t^2}

    Thus the midpoint of the ladder has the coordinates M\left(\dfrac t2\ ,\ \dfrac12 \sqrt{36-t^2} \right)

    Calculate the locus of all M by solving the system of equations:

    x = \dfrac t2~\implies~t = 2x

    y = \dfrac12 \sqrt{36-t^2}~\implies~4y^2=36-4x^2~\implies~x^2+y^2=9

    That means the locus is a circle around the origin with a radius of 3.
    Attached Thumbnails Attached Thumbnails [SOLVED] Locus-leiter_anwand.png  
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