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 Share with your friends Share 17 Tanveer Sofi answered this 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=-1⇒bakabh×ba=1⇒a3k=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=1⇒hc2a3=1⇒h=a3c2 ⇒a=hc213 As, a2+b2=c2So, k=b3a3h=b3a3a3c2=b3c2⇒b=kc213Putting the values of a and b in i, we gethc223+kc223=c2⇒h23+k23=c23Therefore, the required locus of Q isx23+y23=c23. 56 View Full Answer