Two ends A and B of a straight line segment of constant length c slide upon a fixed rectangular axes OX and OY respectively. If the rectangle OAPB is completed. Then find locus of the foot of the perpendicular drawn from P to AB



Let the coordinates of A  and  B be a,0  and  0,b respectively. As the rod slides the values of a  and  b change, so a  and  b  are variables.Also, The point P is formed by completing the rectangle OAPB, so the coodinates of P are a,b.Let Q h,k be the foot of perpendicular of perpendicular from P on AB.As, the length of the rod is constant c, thena2+b2=c2        ........iSlope of  line  AB=-baSlope of line AQ=-ka-h Slope of line QB=-b-khAs, A,Q, B lie on the same line, therefore-ba=-ka-h=-b-kh a-h=abk   and   b-k=bahAlso, slope of PQ=b-ka-hNow, PQ  is perpendicular to AB, thenb-ka-h×-ba=-1bakabh×ba=1a3k=b3h          .........iiAlso, using the intercept form, the equation of line AB isxa+yb=1As, Qh,k  lies on AB, thereforeha+kb=1           ...........iiiPutting k=b3a3h  from ii  in equation iii, we getha2+b2a3=1hc2a3=1h=a3c2 a=hc213               As, a2+b2=c2So, k=b3a3h=b3a3a3c2=b3c2b=kc213Putting the values of  a  and b  in  i, we gethc223+kc223=c2h23+k23=c23Therefore, the required locus of Q isx23+y23=c23.

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