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#?

Question

Allison Allen · Accepted Answer

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

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:

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]}

Violet Aldrich · Answer

undefined The locus of point P can be found by understanding the geometric constraints involved. Let's denote the coordinates of point P as (x, y).

The area of the quadrilateral formed by the two lines y = 0 and y = mx and the perpendicular from P to these lines remains constant. Let A be this constant area.

The perpendicular from P to the line y = 0 will intersect the x-axis at a point with coordinates (x, 0), and the perpendicular from P to the line y = mx will intersect the line y = mx at a point with coordinates (x, mx). The distance between these two points will give the height of the quadrilateral, which is |mx|.

The length of the base of the quadrilateral (along the x-axis) is |x|.

Therefore, the area of the quadrilateral is given by the formula: Area = (1/2) * |mx| * |x| = (1/2) * |mx^2|.

Since this area is constant, we have |mx^2| = 2A.

This implies that |x^2| = 2A / |m|.

Hence, the locus of P is given by the equation |x^2| = 2A / |m|.

This represents a pair of curves, one in the first and third quadrants and the other in the second and fourth quadrants, symmetric about the y-axis.