A point #P# moves between lines #y=0# and #y=mx# so that the area of quadrilateral formed by the two lines and perpendicular from #P# on these lines remains constant. Find the equation of locus of #P#?

1 Answer
Feb 8, 2018

The equation is of the locus is of type #my^2-mx^2+2xy=k# and it is a hyperbola.

Explanation:

Let us consider that equation of line #OA# is #y=0# (#x#-axis) and that of line #OB# is #y=mx# and #O# is origin. Let the coordinates of #P# be #(x,y)#. The diagram appears as follows:

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Point #P(x,y)# moves so that area of quarilateral #OMPN# is some constant.

Now it is evident that area of #DEltaPMO=1/2xy#. For area of #DeltaPNO#, and #OP=sqrt(x^2+y^2)#. Let us work out #PN# and #ON#. It is evident that #PN=|(y-mx)/sqrt(1+m^2)|#

and hence #ON^2=x^2+y^2-(y-mx)^2/(1+m^2)#

= #(x^2+y^2+m^2x^2+m^2y^2-y^2-m^2x^2+2mxy)/(1+m^2)#

= #(x^2+m^2y^2+2mxy)/(1+m^2)=(x+my)^2/(1+m^2)#

and #OP=|(x+my)/sqrt(1+m^2)|#

Hence area of #DeltaPNO# is

#1/2((y-mx)(x+my))/(1+m^2)=1/2(my^2-mx^2+xy-m^2xy)/(1+m^2)#

and area of quadrilateral is

#1/2xy+1/2(my^2-mx^2+xy-m^2xy)/(1+m^2)#

= #1/2((xy+xym^2+my^2-mx^2+xy-m^2xy)/(1+m^2))#

= #1/2((xy+my^2-mx^2+xy)/(1+m^2))#

Hence equation of #P(x,y)# would be

#my^2-mx^2+2xy=k#, where #k# is a constant.

This is the equation of a hyperbola.

Below is shown the graph for #m=2# and #k=20#

graph{y^2-x^2+xy=10 [-20, 20, -10, 10]}