Friday, December 18, 2015

Locus Problem - Feet of Perpendicular

Locus Problem : A ladder of length L is sliding over the floor. Find the locus of point D such that D is the foot of perpendicular dropped from point C such that OACB is a rectangle.

Solution: Let A (a,0) be the feet of the ladder and B (0,b) be the top of the ladder rest against the wall . O is the intersection point of floor and wall. OACB is a rectangle. D is the feet of the perpendicular drawn from point C onto ladder AB . As the feet of the ladder slides, the point D travels on a path called asteroid whose equation is given by x(2/3)+y(2/3)=L(2/3).

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of point D of the ladder. You can also select the ‘Show Locus’ button to see the locus and its equation.

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