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The ends of a rod of length l move on tw...

The ends of a rod of length l move on two mutually perpendicular lines. Find the locus of the point on the rod which divides it in the ratio `1:2`.

Text Solution

Verified by Experts

The correct Answer is:
`36x^(2)+9^(2)=4l^(2) or 9x^(2)+36y^(2)=4l^(2)`
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A rod of length l slides with its ends on two perpendicular lines. Find the locus of its midpoint.

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Knowledge Check

  • A straight line segment AB of length 'a' moves with its ends on the axes. Then the locus of the point P which divides the line in the ratio 1:2 is

    A
    `9(x^2+y^2)=4a^2`
    B
    `9(x^2+4y^2)=4a^2`
    C
    `9(y^2+4x^2)=4a^2`
    D
    `9x^2+4y^2=a^2`

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    A line of fixed length a+b moves so that its ends are always on two fixed perpendicular straight lines. Then the locus of the point which divides this line into portions of length aa n db is (a) an ellipse (b) parabola (c) straight line (d) none of these

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    A is a point on the circle x^(2)+y^(2)=36 . Find the locus of the point P which divides the ordinate bar (AN) internally in the ratio 2:1 .

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